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https://byjus.com/maths/class-9-maths-chapter-1-number-system-mcqs/?replytocom=308812 | 1,638,126,258,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358570.48/warc/CC-MAIN-20211128164634-20211128194634-00567.warc.gz | 221,401,467 | 156,243 | # Class 9 Maths Chapter 1 Number System MCQs
Class 9 Maths Chapter 1 Number System MCQs are available here with solutions. The MCQs are prepared as per the CBSE syllabus (2021-2022) and NCERT curriculum. These objective questions are given online, as per the latest exam pattern. Also, we have provided detailed explanations for some questions to help students understand the concept very well. Solving the chapter-wise MCQs for 9th Standard Maths subject will help students to boost their problem-solving skills and confidence. Also, check Important Questions for Class 9 Maths here.
Download the below PDF to get more MCQs on Class 9 Maths Chapter 1 Number System.
## MCQs on Class 9 Maths Chapter 1 Number System
Check the below multiple choice questions for 9th Class Maths chapter 1-Number system. All MCQs have four options, out of which only one is correct. Students have to choose the correct option and check the answer with the provided one.
1.) Can we write 0 in the form of p/q?
a. Yes
b. No
c. Cannot be explained
d. None of the above
Explanation: 0 is a rational number and hence it can be written in the form of p/q.
Example: 0/4 = 0
2.) The three rational numbers between 3 and 4 are:
a. 5/2, 6/2, 7/2
b. 13/4, 14/4, 15/4
c. 12/7, 13/7, 14/7
d.11/4, 12/4, 13/4
Explanation: There are many rational numbers between 3 and 4
To find 3 rational numbers, we need to multiply and divide both the numbers by 3+1 = 4
Hence, 3 x (4/4) = 12/4 and 4 x (4/4) = 16/4
Thus, three rational numbers between 12/4 and 16/4 are 13/4, 14/4 and 15/4.
3.) In between any two numbers, there are:
a. Only one rational number
b. Two rational numbers
c. Infinite rational numbers
d. No rational number
Explanation: Take the reference from question number 2 explained above.
4.) Every rational number is:
a. Whole number
b. Natural number
c. Integer
d. Real number
5.) √9 is __________ number.
a. A rational
b. An irrational
c. Neither rational nor irrational
d. None of the above
Explanation: √9 = 3
Hence, √9 is a rational number.
6.) Which of the following is an irrational number?
a. √16
b. √(12/3)
c. √12
d. √100
Explanation: √12 cannot be simplified to a rational number.
7.) 3√6 + 4√6 is equal to:
a. 6√6
b. 7√6
c. 4√12
d. 7√12
Explanation: 3√6 + 4√6 = (3 + 4)√6 = 7√6
8.) √6 x √27 is equal to:
a. 9√2
b. 3√3
c. 2√2
d. 9√3
Explanation: √6 x √27 = √(6 x 27) = √(2 x 3 x 3 x 3 x 3) = (3 x 3)√2 = 9√2
9.) Which of the following is equal to x3?
a. x– x3
b. x6.x3
c. x6/x3
d. (x6)3
Explanation: x6/x3 = x6 – 3 = x3
10.) Which of the following is an irrational number?
a. √23
b. √225
c. 0.3796
d. 7.478478
Explanation: √23 = 4.79583152331…
Since the decimal expansion of the number is non-terminating non-recurring. Hence, it is an irrational number.
11.) Which of the following is an irrational number?
a. 0.14
b. $0.14\overline{16}$
c. $0.\overline{1416}$
d. 0.4014001400014…
Explanation: 0.4014001400014…is an irrational number as it is non-terminating and non-repeating.
12.) 2√3+√3 =
a. 6
b. 2√6
c. 3√3
d. 4√6
Explanation: 2√3+√3 = (2+1)√3= 3√3.
13.) Find the value of $\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}$, if √2 = 1.4142.
a. 5.8282
b. 0.1718
c. 0.4142
d. 2.4142
Explanation:
$\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}$ = $\sqrt{\frac{1.4142-1}{1.4142+1}}$
=$\sqrt{\frac{0.4142}{2.4142}}$
= 0.4142
14.) The number obtained on rationalising the denominator of $\frac{1}{\sqrt{7}-2}$ is
a. (√7+2)/3
b. (√7-2)/3
c. (√7+2)/5
d. (√7+2)/45
Explanation:
$\frac{1}{\sqrt{7}-2} = \frac{1}{\sqrt{7}-2}\times \frac{\sqrt{7}+2}{\sqrt{7}+2} = \frac{\sqrt{7}+2}{(\sqrt{7})^{2}-(2)^{2}} = \frac{\sqrt{7}+2}{3}$
15.) Which of the following is rational?
a. 4/0
b. 0/4
c. √3
d. π
Explanation: 0/4 is a rational number that is equal to 0. Whereas π and √3 are irrational numbers and 4/0 is meaningless.
16.) The irrational number between 2 and 2.5 is
a. √11
b. √5
c. √22.5
d. √12.5
Explanation: The irrational number between 2 and 2.5 is √5 because the approximate value of √5 is 2. 23606…
17.) The value of √10 times √15 is equal to
a. 5√6
b. √25
c. 10√5
d. √5
Explanation: √10×√15 =(√2.√5)×(√3. √5) = 5√6.
18.) The decimal representation of the rational number is
a. Always terminating
b. Either terminating or repeating
c. Either terminating or non-repeating
d. Neither terminating nor repeating
Explanation: As per the definition of rational number, its decimal representation is either terminating or repeating.
19.) Which of the following is a rational number?
a. 0
b. 2√3
c. 2+√3
d. π
Explanation: 0 is a rational number, and it can be written as 0/1 or 0/2 or 0/3 etc. Whereas 2√3, 2+√3, and π are irrational numbers.
20.) Which of the following is an irrational number?
a. √(4/9)
b. √12/√3
c. √7
d. √81
Explanation: √7 is an irrational number. Because other options given are simplified into a rational number.
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1. ## How to set Temperature as load in Thermal Analysis?
Hi Rahul, what is TREF in MAT1 card. I am simulating exhaust manifold and having temperature distribution over all nodes. I have run the linear static analysis but stresses are very very high >500MPa. When I am useing Tref =400 deg C, then stresses are low. My average temp over exhaust manifold nodes is 500deg C. Please comment. Thanks in advance. Regards, Dheeraj
2. ## postprocessing of abaqus result file in hyperview
Hi, I have gone thru the hyperview post processing document where corner data & averaging method is explained for shell and solid element. Those explanations are bit confusing. Please clarify if I have done static analysis of Floor assembly in Abaqus & have .odb result file and want to do post processing in Hyperview. Then in Abaqus CAE, I am quoting the stress as Simple average (75%). To correlate same result,which option is best tin Hyperview- 1) Corner data should be on or off? 2)Averaging method should be simple,NONE or Advanced? 3) which combination of corner data & averaging method. I tried with all options and I got 20% deviation in hyperview. Thanks in Advance. Regards Dheeraj Pandey
3. ## rms stress and psd elemental stress
Hi, what will be the unit & calculation method of PSD stress . If I quote von mises stress under PSD stress then what result I can interpret Thanks in advance.
4. ## FRF analysis
Thanks for sharing the paper. But that was done with defined load as excitation. My query is different. What excitation (g value) should I use in CAE for FRF analysis of an overhang part - Example If I found acceleration peak of 2 G at 1500 RPM in engine sweep up data ( acceleration vs RPM -idle to full throttle). My engine is 4 cyl - 4 stroke then engine resonance freq is (1500/60)*2 = 50Hz then when I again run the engine at 1500RPM (50 Hz), and get FFT data - Acceleration vs Frequencies. So, at 1500 RPM, there are various acceleration peaks at 50 Hz and its harmonic. But its acceleration values are 0.7 G or 0.5 G. So, for FRF analysis in CAE, which G value I should take - 2 G or 0.7 G?? Thanks in advance
5. ## PSD analysis
Thanks for the information. At the time of postprocessing, which output is more useful to determine the life cycle. PSD or RMS??
6. ## FRF analysis
Hi, I am performing FRF analysis of an overhang part mounted on engine. From testing, I requested following vibration data at mounting and at tip- 1) G vs RPM (sweep up) - from this I got to know at which RPM, G value is maximum due to resonance. 2)FFT data (G vs Freq) at that single RPM where G was maximum while sweep up. Now, my query is ...from FFt data, my maximum G value at particular frequency is less than 1G . But from sweep up data (G vs RPM), if I check G value at that RPM , then it is more than 3G. So, Now For FRF analysis in CAE, which G value I should take for excitation input (from FFT or from sweep up). How to correlate max. displacement and acceleration plotted at tip in CAE with testing. Thanks in advance.
7. ## PSD analysis
Hi, I am currently performing random vibrartion analysis, where I have sweep up PSD data for muffler. I have some query mention below: 1)From testing, I got PSD data (G^2/Hz vs Freq) . Do I have to change it into { (mm/sec2)^2 vs freq}. 2)when I am giving the input in (G^2/Hz vs Freq) and FRF excitation as 1 , then PSD stress is very very less e-7. 3)Plz suggest the appropriate methodology step by step for providing - input (in which form) and post-processing method to interpret results (which stress is useful). Thanks in Advance
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# HOW TO DESTROY READING COMPREHENSION PASSAGES BY RHYME
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05 Jun 2006, 19:14
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HOW TO DESTROY READING COMPREHENSION PASSAGES BY RHYME
There is a magic bullet for RC passages.
What? There's a magic bullet? Surely you lie rhyme!
I do not. Here it is: Don't read the passage.
Suprised? It works. The primary purpose of the RC passages are not to test your knowledge of grammar or theories or anything like that - they test your ability to RETAIN INFORMATION. Problem is, you've got a few minutes to read a passage on some of the most boring crap ever, and you somehow have to remember it? It can't be done.
So, how do you beat the RC down to it's knees and kick it in the groin?
You only read parts of it.
Lets try this with a long passage attached. (DONT READ IT YET!!)
Oh good god, thats atrociously long! So what do you do?
Step 1: Read the first paragraph and rewrite the key points. Rewrite in your own words.
Step 2: Read the first sentence of each subsequent paragraph. Rewrite in your own words.
Step 2a: SKIM the paragraph looking for key words - names, dates, key words. Write these down underneath the key sentence you wrote for each paragraph.
NOW TRY THIS ON THE ATTACHED PASSAGE BEFORE YOU READ MY NOTES BELOW
What did your notes look like?
My notes might look something like this:
Black Death severe epidemic, ravaged 14th cent Europe. Intrigured scholars since Gasquet 1893 study. Gasquet contends epidemic intensified political / religious upheaval that ended middle ages. Later, Coulton agreed but oddly attributed a good thing to the BD - propersity as a result of less competition for food, shelter and crap.
1930s, Evgeny Kosminksy claimed epedemic as not a key player.
World War, Marxist, fedualism
Role of BD also challenged in other way.
Twigg, Sherwburry, trade ship, havoc, bubonic, nile, 1912
Although Twigg cites conditions needed for BD, he ignores too much and is faulty in his logic.
Speculation, fault, trade ship, rodents, animals, europe
..........................
Now I've boiled the entire thing down to a few sentences. Try re-reading the first sentences now if you are confused about the point of the passage. Whats the passage saying? DB was bad, lots of people have studied it, one guy argues it helped end the middle ages, some other guy said it helped foster prosperity, someone else argued against that, some other guys cahllenged it too, some guy named Twig is wrong. Ok, so the authors talking about the DB, and specifically some different theories about it. No problem.
What happens though when you get the question:
"Which of the following statements is most compatible with Kosminksy's approach to history as it is presented in the passage?"
Easy. Find Kosminsky in your notes. Oh, there he is, in the second paragraph. Ok, now go look back at the second paragraph. Find his name. Read ONE sentence around his name. If you don't see the answer, read TWO sentences. If it's not in either of those sentences, see if his name comes up somewhere else in the passage. The answers to the specific questions become REALLY REALLY EASY if you use this method. Why? The GMAT LOVES to test your ability to remember the impossible.
What are the answer choices for this question?
Quote:
(a) The middle ages were ended primarily by the religious and political upheaval in fourteen century europe
(b) The economic consequences of the BD included increased competition for food, shelter and work.
(c) European history cannot be studied in isolation from that of hte rest of hte world
(d) The number of deaths in the fourtheenth century has been exaggerated
(e) The significance of the black death is best explained within the context of evolving economic systems.
Do you see the GMAT's trap? They do this ALL the time with specific questions like this one. "OOOH OOH I REMEMBER READING ABOUT HOW THE MIDDLE AGES WERE ENDED BY RELIGIOUS UPHEAVAL... ILL PICK THAT." Or, maybe you don't remember that and you pick B becuase IT LOOKS FAMILIAR AND YOU REMEMBER IT. How many names came up in this passage? A half dozen? Evgency, Coulton, Gasquet, Twigg, Shrewsbury! The gmat is trying to trick you to do one of two things - either (A) pick based on what you remember or (B) worse, make you go back and re-read half the Oops passage.
You will do neither of these.
Now, go back and read only two sentences around the word Evgeny Kosminsky. Do you see the answer? There's only one possible answer that even COMES CLOSE. Lets say you have no FRICKING CLUE what the hell Kosminsky is trying to say, even if you HAVE NO Oops CLUE, there's only one option that has a very similar word to those two sentences. "economic" and "economically". How easy did that become?
Now what if they asked you a general question?
Quote:
The passage is primarily concerned with:
A) Demonstrating the relationship with the bubonic plague and the black death
B) Interpreting historical and scientific works on the black death
C) Employing the black death as a case study of disease transmission in medieval Europe
D) Presenting aspects of past and current debate on teh historical importance of the black death
e) Analyzing differences between capitalist and marxist interpretations of the historical signficance of the black death
No !@(*!(#@ problem. Remember how you broke down the passage in to a few sentences? What did it say? Did you write bubonic plague anywhere? No, not in any of your key sentences. Eliminate A. Is there any mention of case studies anywhere? No not really, so eliminate C. Did you write down anything about capitalists? No, eliminate E. Ok, so you are down to B and D. Look back at your sentences - is the author interpreting things for you or just telling you that there are different views? In other words, is he interpreting or presenting? He's presenting. Answer is D.
Did you get both these questions right? Hopefully you did. Did you notice how you never actually read the !@(#!(@ passage?. Cool eh?
I really hope that made some sense. In my mind, this is the fullproof way of DESTROYING the RC on the GMAT. You can obliterate it if you take the time to do this stuff. Oh and don't forget, its much faster to read twelve sentences than to read 70.
Someone pointed out a stickied verbal thread called "USeful verbal documents" or something like that. In there, it says this about RC:
Quote:
Try to read the whole text of the passage once, if possible. Many people think you should just skim the passage or read the first lines of every paragraph, and not to read the passage. We believe this is an error: if you misunderstand the main idea of the passage, you will certainly get at least some of the questions wrong. Give the passage one good read, taking no more than 3 minutes to read all of the text. Do not read the passage more than once – that wastes too much time. If you have not understood it completely, try to answer the questions anyway.
A few comments. First, I'm not advocating you skim the passage. I'm advocating you read the entire first paragraph and the first sentence of each subsequent one, and then skim. What I find shocking in the advice above are two things:
1) "If you don't the main idea of the passage you will certainly get at least some of the questions wrong."
Not necessarily. This is only true if you get a bunch of general questions, but you are EQUALLY likely to get a bunch of specific questions - where your understanding of the whole passage is not important.
2) "Do not read the passage more than once – that wastes too much time. If you have not understood it completely, try to answer the questions anyway."
Good Lord. Who came up with this strategy? Read the whole thing once, if you don't understand it, try anyway! You want to talk about a sure fire way of NOT getting things right? Think about it... the whole POINT of RC is to test your ability to retain information, the whole POINT of the questions is to try and force you to go back multiple times and re-read sections again and again. According to the strategy posted in the word doc in that thread, you should just read it once and then "do your best"? Sorry, but this has got to be some of the worst advice I could imagine.
Reading the whole passage once will do a few things:
(1) It will take more time than my method, AND you won't have any notes at the end!
(2) You will GUARANTEE confusion. There is a reason the GMAT picks dry scientific passages and not passages from some Tom Clancy novel. (Even though those suck too). It's because they are PACKED with information, often TERTIARY information - it's meant to be hard to digest this stuff. On top of it, they suggest 3 minutes to read AND understand the text?
The advice they give sounds familiar. It sounds like Kaplan. Read the whole thing! Then take notes! Then answer questions! HAY GUYZ, ITS 75 MINUTES YOU KNOW? It's crazy advice.
Theres one more thing I want to say about RC.
You know those questions about "The author infers....." or "It can be inferred...." ?
I'm going to try and find an example of this tonight if I can, but when they say that, they really DONT want you to infer much. They really just want you to find what the author said. I can't explain this very well without an example, but I will look for one. If anyone knows of one in the book somewhere, just PM me it or give me a page/probl # or post it here.
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06 Jun 2006, 08:28
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But your approach is similar to Kaplan's. You are suggesting that we paraphrase the passages in our own words, so we will need to read the passage anyway!!!
The problem with paraphrasing in your own words is that it takes time! I can't afford to paraphrase a 500 word passage in 50 words in less than 5-6 mins and it hardly leaves me any time for answering the questions.
Secondly, what do you mean by skimming? If I have to understand what I'm reading, there's no difference between skimming and reading. Secondly, if I don't understand anything and still read it anyways, what's the point of skimming? I very well can leave the text alone altogether!
Please elaborate a bit on the approach for skimming as I can't seem to grasp the same.
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I am also going through same issue! If I don't read passage properly can not answer many questions; Off course, reading passage properly and still making mistake is different beast all together! Though, I found that skimming and partially reading, in the end, may take more time, partially due to re-reading, and partially due to lack of confidence! Though, I believe both techniques has its plus and minuses.
Consider the time when you finish answering first question for around 400 word paragraph.
Style 1: Reading properly: Read paragraph 4-5 min, answer first question 1 min so 5-6 min for first question
Style 2: Reading first paragraph and first sentence successive paragraph.. 2-3 min answering first question 1-2 min.. So first question could be answered in 3 to 5 min..
Depending on paragraph, one of the techniques may work better than other, but when you are really rushing and willing to accept lower hit rate to speed up second techniques may be better as you can cut your losses by skipping question that you can not answer in a min or so.. So take a hit and move on!
So I have decided to use both techniques! Read properly and skim! And planning to practice and dynamically changing as per situation.
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alfa_beta01 wrote:
But your approach is similar to Kaplan's. You are suggesting that we paraphrase the passages in our own words, so we will need to read the passage anyway!!!
No, thats not what I'm saying - I think you the issue is how you understand skimming.... . I'm saying PARAPHRASE THE FIRST PARAGRAPH and THE FIRST SENTENCE OF EACH PARAGRAPH. Thats it. Not the entire passage. DON'T EVEN READ the other paragraphs, just skim and write down names or dates in case they come up in the exam. If you are good at skimming text to look for names - you can do what I do, dont even bother skimming the paragraphs. Just read the first paragraph and the first sentence of each paragraph.
Quote:
Secondly, what do you mean by skimming? If I have to understand what I'm reading, there's no difference between skimming and reading. Secondly, if I don't understand anything and still read it anyways, what's the point of skimming? I very well can leave the text alone altogether!
By skim, all I mean is GLANCE. I don't read sentences.
For example: (dont read it yet, read my notes first, then try it)
If the second paragraph said:
"Though the basic concept of the strip is straightforward, Herriman always found ways to tweak the formula. Sometimes, Ignatz's plans to surreptitiously lob a brick at Krazy's head succeed; other times Officer Pupp outsmarts the wily mouse and imprisons him. The interventions of Coconino County's other anthropomorphic animal residents, and even forces of nature, occasionally change the dynamic in unexpected ways. Other strips have Krazy's simple-minded or gnomic pronouncements irritating the mouse so much that he goes to seek out a brick in the final panel."
Paraphrase first sentence: (I'll time myself)
"Strip straightforward, but author found ways to tweak. "
That took me about 20 seconds.
NOW, DO NOT READ THE REST OF THE PARAGRAPH. GLANCE OVER IT, JUST LOOKING FOR KEY WORDS. NAMES, DATES, THINGS OF THE SORT. (I'll explain why in a minute)
All I would write is: (I'll time myself again)
Herriman, Ignatz, Krazy, Officer Pupp, mouse, Coconino, residents, Krazy.
I did that in about 15 seconds. . I know that sounds crazy, but try it - just glance over it looking for names, dates, or otherwise things that might appear important - anything in capital letters for example. This includes scientific names. So if you see something like "proto-plasmic neurons" in the middle of a passage, scribble it down.
I haven't read the above paragraph, and to be honest I have no idea what it says.
In order to understand why this works you have to understand the two types of RC questions that come up.
Type 1 is general. These are questions that ask things like "The author is most concerned with" or "the primary purpose of hte passage is" or "the author would most agree with..."
IN ORDER TO ANSWER THESE YOU ONLY NEED TO KNOW THE GENERAL POINT OF THE PASSAGE. THIS POINT IS ALWAYS MADE DURING THE FIRST PARAGRAPH AND TOPIC SENTENCES (FIRST SENTENCE) OF EACH SUBSEQUENT PARAGRAPH. THATS IT. YOU DO NOT HAVE TO KNOW ANYTHING ABOUT THE DETAILS BEHIND THE GENERAL POINT. IN FACT, USUALLY, KNOWING THESE DETAILS THROWS YOU OFF BECAUSE YOU'LL SEE AN ANSWER THAT LOOKS REALLY SIMILAR TO SOMETHING YOU JUST READ.
Thats type 1. Type 2 is the SPECIFIC QUESTION. These are things like "Dr. Neilsen agrees that ...." or "In 1914, the primary differences between colonial europe and the americas were... " or "The author states what about proto plasmic neurons?"
In every single case, these questions will boil down to a sentence, or at most two sentences, SOMEWHERE in the passage.... thats it. So, if you scribbled down key words, it should take you a few seconds to figure out what paragraph the answer is in and another few seconds to read the setnence it's in. The answer is always there - plus or minus one sentence.
The key to understanding this methodology is: you are not trying to understand the passage. You only need understand the first paragraph, and first sentence of each piece.
We'll do one more together.
I'll time myself and try to write down details of what I read and did not read.
It's attached at the bottom.
First paragraph:
Time taken to read and take notes: 1 minute 20 seconds
My notes: Solar ponds circulation incomplete + high salt concentration that increase w/ depth. This traps heat. Low water traps heat, higher water insulate. Heat thus retained at depth.
Second paragraph:
Time taken to read first sentence and paraphrase: 27 seconds.
My notes: Artificial pond made in dead sea to test its ability to convert heat to electricity.
Second paragraph SKIMMING:
Time taken to SKIM and write down key words: 17 seconds.
My notes: water. solar ponds, chemicals, penetration, algae
(NOTE how quickly that was done. SKIM SKIM SKIM)
Third paragraph:
Time taken to read first sentence and paraphrase: 9 seconds.
My notes: Algicide proposed to control algea.
Third paragraph SKIMMING: 16 seconds.
My Notes: Dead sea, chemcials, lucrative, tourist, contaminated.
Fourth paragraph first sentence: 15 seconds.
My notes: Recent exp more promising for controlling algea
Fourth paragraph skim: 24 seconds.
My notes: repress, distortion, bouyancy, storage layer, destroyed , evaporation, diluted, algea.
Total time taken: 3 MINUTES 20 SECONDS
Thats pretty decent, maybe even a bit slow - but notice how quickly I ate up the remaining paragraphs. I skimmed paragraphs 3 and 4 in under a minute....
Its not about READING the paragraphs. Just skimming. You may have noticed that my skim notes are out of order - words that come later are first... the reason this is the case is because I let my eyes see a word, write it down, if it then sees another word that I didnt see, even if its before, i write it down. I know that finding the word again will take me a few seconds - if you can write them in order, all the better. I found that I can be much faster if I just glance and write - even if its out of order.
How did I do it so fast? I have not read a single sentence (other than the first sentence) in paragraph 2,3 or 4. I have no idea what details are in there - but I do know what they generally talk about BECAUSE I WROTE DOWN THE FIRST SENTENCE, WHICH ALWAYS INTRODUCES THE PARAGRAPH'S TOPIC. That's all I need to know to answer a question like:
"What is the author primarily concerned with?"
Well, without even looking at answer choices, I know he's primarily concerned with salt water ponds and their suitability to generate electricity.
I know this because he introduces them in the first paragraph, mentions an expirement in #2, cites a possible solution to some problem in #3, and then says theres a better solution in paragraph #4. I know all this just by reading hte first setnence of each paragraph. I do NOT know WHY or WHAT the better solution is - but if I get asked, I know where to look. (It'll be somewhere in paragraph #4)
QUESTION: The primary purpose of the passage is to:
(a) discuss ways of solving a problem that threatens to limit the usefullness of an energy source.
(b) explain the mechanisms by which solar heat may be converted into energy
(c) detail the processes by which algae colonize highly saline bodies of water
(d) report the results of an experiment designed to clean contaminated bodies of water
(e) describe the unique properties of solar ponds in the dead sea.
Can you see which one it is?
Look again at the first sentences and the first paragraph.
(1) Solar ponds retain heat
(2) Artificial pond made in dead sea to test its ability to convert heat to electricity.
(3) Algicide proposed to control algea.
(4) Recent exp more promising for controlling algea
What story is this telling? There are some ponds, there's one in the dead sea, theres some problem with it, algicide is one solution, but there is a better one. Look again at the answer choices.
Do you see it?
(b) explain the mechanisms by which solar heat may be converted into energy ------- Well no, that doesnt really jibe with the topic sentences... He mentions it, but the topic sentences aren't describing a mechanism, they decribe a problem.
(c) detail the processes by which algae colonize highly saline bodies of water ---- No, again, theres not much discussion of processes in the topic sentences. The guys talking about a problem.
(d) report the results of an experiment designed to clean contaminated bodies of water --- Possibly, but this is really only part of the passage probably. I dont know for sure, because I haven't read it, but again if you look back at the notes, he's not so much reporting as much as suggesting something.
(e) describe the unique properties of solar ponds in the dead sea. - Well, definetly not. For starters, i dont remember it saying solar ponds in the dead sea were the only ones that existed, plus, again, the word Dead Sea doesnt come up much in the topic sentences.
(a) discuss ways of solving a problem that threatens to limit the usefullness of an energy source. - Hmm ok, that makes sense. There is a pond, there is an experiment to check it as an energy source, there are some solutions to that problem, but one solution is better. Yea, thats exactly what he's saying.
Now say you got a specific question:
"According to the passage, the growth of algae was considered a threat to the sucess of the artificial pond near the dead sea beacuse the algae..."
(a) produce excess oxygen that lowers water temperature
(b) restrict the circulation of the pond
(c) enable heat to escape through the upper level of the pound
(d) prevent light from penetrating to the lowest level of hte pond
(e) prevent accurate measurement of the heat collected in the pond.
How do you solve this given that YOU HAVENT EVEN READ THE PASSAGE?
Just find where Algae and Dead sea come up. Dead sea comes up first in the second paragraph - we know this because I wrote it in my SKIM NOTES for the second passage.
Did the word algae come up?
Lets go look at what I wrote as my SKIM notes for paragraph two.
My notes: water. solar ponds, chemicals, penetration, algae
Yea, there it is. If your notes are lucky, you might notice that I wrote down the words algae and penetration... gives you a pretty huge hint what it might be, but thats kind of dumb luck... so lets pretend I didnt write that down.
How do we find the answer?
Find the sentence with algae in the second paragraph, because that's where we first saw Dead Sea.
"An immediate threat to the sucess of the venture was the growth of algae".
Ok, that doesnt give us the answer, what about the sentence right next to it.
"Water in solar ponds must be kept maximally transparent to allow penetration of light to the deep storage area."
Ok, look back at the answer choices.
The answer should jump out at you now. It's D.
It's not A because theres NO MENTION of oxygen.
Its not B becuase theres NO MENTION of circulation.
It's not C becuase theres NO MENTION of heat
Its not E because theres NO MENTION of measurement.
The only answer that even has words that match words in that sentence is D.
How long did it take you to find them? 10, maybe 20 seconds. Tops. 20 more seconds to read them. You've answered the question in under a minute.
How long do you think it would have taken to find that answer otherwise? I mean, all the other options SOUND reasonable right? Circulation was mentioned in paragraph 1, heat definetly came up somehwere, and there might even be a mention of oxygen somewhere I missed.
the point is this:
YOU DO NOT HAVE TO READ THE PARAGRAPHS TO GET THESE RIGHT. ONLY READ THE FIRST PARAGRAPH AND FIRST SENTENCE. SKIM THE REST. AND BY SKIM, I MEAN 20 SECONDS, KEY WORDS, DONT EVEN READ THE SENTENCES, JUST WRITE DOWN SCRIBBLES.
With time, you can get really really fast at it.
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06 Jun 2006, 20:52
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Thanx Rhyme for your wonderful post.I think it makes sense to approach the passages your way.
I'll give it a try and let's see if it works for me....
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07 Jun 2006, 07:39
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Thanks for the detailed explanation Rhyme; I will give it a shot and see how it goes. By the way any idea how long it might take to master this approach?
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07 Jun 2006, 15:12
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alfa_beta01 wrote:
Thanks for the detailed explanation Rhyme; I will give it a shot and see how it goes. By the way any idea how long it might take to master this approach?
I took to it quickly... id say after about 6 or 7 RC's (passages, not questions) I had it down.
At first you will be slow...but dont fret if it takes you 6 minutes at first. Give it a good try. Take an afternoon, pick out a few RC's and try it.... don't just give up after two. After you've tried doing three or four passages (answer the questions too - taking notes is only half the equation - you stil have to practice finding the answer from your notes) - try timing yourself. It will probably take you five or six minutes. Thats OK. After you've done that, do one or two more, then time how long it takes you to get through the passage - and seperately time how long each question takes you. As you get the hang of it, you'll realize that the questions themselves can be solved pretty darn fast.
You will also notice that I wrote down some verbs, not just dates or names. I forgot to mention this in the above.... I write down words that seem like key words as well ... important looking verbs (so not things like "is" or "are" or simple things - verbs with basically little meaning) but verbs like "penetrate" or destory, etc.
If you want to compare notes / approach, post an RC here and I'll be happy to try it.
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08 Jun 2006, 00:18
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Thanks for the help Rhyme; really appreciate it. Will bug you in case I have more queries or require help
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08 Jun 2006, 13:44
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Dude - you are awesome! awesome! I will give these kungfu techniques a try and see what happens.
Thanks
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08 Jun 2006, 15:33
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Let me know how they work for you.
Are my explanations clear?
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08 Jun 2006, 16:02
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rhyme wrote:
Let me know how they work for you.
Are my explanations clear?
They are clear and I tried 3 passages using the technique but still need to get used to it. Maybe will give it a shot over the weekend and see how it goes.
Right now I am using up a lot of time hunting for key words and writing them down. And worse, since I am in a hurry, I tend to miss out on a few keywords and have to invariably read the 'skimmed' para(s) again. I guess I will get better with practice.
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11 Jun 2006, 00:11
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writing notes on the pad takes a lot of time..
because most of the time if i write quickly, i dont understand my handwriting......
so what i have done in the past is something similar to yours... but i kind of make a mental map of the comprehension..
is that enough? or should i force myself to make written notes?
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12 Jun 2006, 02:39
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yep my own handwrittings makes me confuse more than paragraph so I am not going to write! though I now started using note of some noun and dates for ex.. para 1, 1962, NMS(Namibu Malibu Srivastava), RC(Robert Chhenee)... so in case I get specific question I can quikly I know where to re-read!
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09 Jul 2006, 17:43
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Ryhme and others-
This strategy works well most of the time for me, but when I have a questions like:
The passage above discusses all of the above EXCEPT:
(a)...(e)
I get stumped, obviously because I have not read the entire paragraph. Most of the times, reading the first lines of paragraph 2 and higher is not enough to answer such questions, and I find my self having to re-read the paragraph JUST FOR this type of question. Do you have a strategy to combat such questions on RC?
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09 Jul 2006, 18:18
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walletless wrote:
Ryhme and others-
This strategy works well most of the time for me, but when I have a questions like:
The passage above discusses all of the above EXCEPT:
(a)...(e)
I get stumped, obviously because I have not read the entire paragraph. Most of the times, reading the first lines of paragraph 2 and higher is not enough to answer such questions, and I find my self having to re-read the paragraph JUST FOR this type of question. Do you have a strategy to combat such questions on RC?
I've found that 90% of the time, these questions come from a paragraph that contains lists. For example, and I dont have GMATPrep installed anymore so I cant say for sure exactly what the passage was, but if you take the exams you'll see it sooner or later...
It says something like:
"People in new england did not believe in religion as part of state, whereas those in england felt strongly that it should. Individuals in england also felt that the lowest class should not be allowed to vote, the exact opposite of traditional england. The new england economy was sparse and heavy in argiculture, a stark contrast to the old country."
I keep an eye out of these lists as I skim.... more often than not these are grouped tightly like this. My notes might say:
NE: not religion and state, E, yes. E, lowest vote, NE, no. E economy spare + argi, E, no.
Sometimes its hard to find them while skimming. What I usually do then is I look at the question options and tackle it backwards. So for example maybe the question says something like:
"The author says all of the following about new england except:"
(A) It's economy was based primary on agricultural products ...
(B) ...
etc
I'll take option 1, pick a key word, i.e. argiculture, and scan the text for it.
When I find it, ill look at options 2 or 3 and see if sentences near the sentence I found with "agriculture" contain anything in the other options. Usually, you'll find that it does. So now its just a question of comparing a half-dozen or so sentences to 5 options on the right...
Sometimes option A is the wrong one - so if I dont find option 1 in the text, ill scan for option 2. The trick is not reading the whole thing but rather picking a key word from the answer choices and scanning for it. You don't have to pick from option 1, if theres a better key word in option 2 or 3 or 4 or 5, scan for that.
Post an example and I'll walk through it if none of htis makes sense...
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09 Jul 2006, 20:43
Thanks Rhyme -
I will post a real example soon. For now, for the example above on solar ponds, if there was a question like:
The author mentions the following about Algae EXCEPT:
a) They interfere with collection of heat in solar ponds
b) They get destroyed in hot water
c) Frequent absorptopm of water by algae cause them to change shape and affects their mobility
d) They affect the salinity of solar ponds
e) They scatter and/or absorb light
How would you approach this? Considering that you have almost spend close to 3.5 minutes reading, skimming and taking notes for the paragraph, and you still do not have enough information to answer this question, it would take a lot of time (atleast a minute, if not more) to find each keyword in the paragraph and read the sentence accompanying it. I seem to be missing something very obvious here.
I completely made up this question based on the example you have above (about solar ponds). I will post a real question from OG or GmatPrep soon too.
Thanks again,
Walletless
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10 Jul 2006, 17:22
Ok, here is a passage from OG:
--------------------------------------------------
Archaeology as a profession faces two major prob-
lems. First, it is the poorest of the poor. Only paltry
sums are available for excavating and even less is avail-
able for publishing the results and preserving the sites
(5) once excavated. Yet archaeologists deal with priceless
objects every day. Second, there is the problem of illegal
excavation, resulting in museum-quality pieces being
sold to the highest bidder.
I would like to make an outrageous suggestion that
(10) would at one stroke provide funds for archaeology and
reduce the amount of illegal digging. I would propose
that scientific archeological expeditions and govern-
mental authorities sell excavated artifacts on the open
market. Such sales would provide substantial funds for
(15) the excavation and preservation of archaeological sites
and the publication of results. At the same time, they
would break the illegal excavator’s grip on the market,
thereby decreasing the inducement to engage in illegal
activities.
(20) You might object that professionals excavate to
acquire knowledge, not money. Moreover, ancient arti-
facts are part of our global cultural heritage, which
should be available for all to appreciate, not sold to the
highest bidder. I agree. Sell nothing that has unique
(25) artistic merit or scientific value. But, you might reply,
everything that comes our of the ground has scientific
value. Here we part company. Theoretically, you may be
correct in claiming that every artifact has potential scien-
tific value. Practically, you are wrong.
(30) I refer to the thousands of pottery vessels and ancient
lamps that are essentially duplicates of one another. In
one small excavation in Cyprus, archaeologists recently
uncovered 2,000 virtually indistinguishable small jugs in
a single courtyard, Even precious royal seal impressions
(35) known as/melekh handles have been found in abun-
dance---more than 4,000 examples so far.
The basements of museums are simply not large
enough to store the artifacts that are likely to be discov-
ered in the future. There is not enough money even to
(40) catalogue the finds; as a result, they cannot be found
again and become as inaccessible as if they had never
been discovered. Indeed, with the help of a computer,
sold artifacts could be more accessible than are the
pieces stored in bulging museum basements. Prior to
(45) sale, each could be photographed and the list of the
purchasers could be maintained on the computer A
purchaser could even be required to agree to return the
piece if it should become needed for scientific purposes.
It would be unrealistic to suggest that illegal digging
(50) would stop if artifacts were sold on the open market.
But the demand for the clandestine product would be
substantially reduced. Who would want an unmarked
pot when another was available whose provenance was
known, and that was dated stratigraphically by the
professional archaeologist who excavated it?
------------------------------------------------------------
Question:
The author implies that all of the following statements about duplicate artifacts are true EXCEPT:
(A) A market for such artifacts already exists.
(B) Such artifacts seldom have scientific value.
(C) There is likely to be a continuing supply of such artifacts.
(D) Museums are well supplied with examples of such artifacts.
(E) Such artifacts frequently exceed in quality those already catalogued in museum collections.
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12 Jul 2006, 18:10
Hi Rhyme -
I am not sure if you missed my previous post. I am just bumping the thread, in case you did not notice.
Thanks,
Walletless
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12 Jul 2006, 20:13
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Yea, I did. I'm drunk at the moment, so I wont try to do it now cause I'll make a mess of it. I'll try to get to your post tomorrow. Sorry for the delay.
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13 Jul 2006, 09:15
rhyme wrote:
Yea, I did. I'm drunk at the moment, so I wont try to do it now cause I'll make a mess of it. I'll try to get to your post tomorrow. Sorry for the delay.
Hehe I COMPLETELY understand that :D Thanks for the update!
13 Jul 2006, 09:15
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Display posts from previous: Sort by | 9,699 | 39,051 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2017-22 | latest | en | 0.903351 |
https://convertoctopus.com/35-5-cubic-inches-to-liters | 1,600,758,785,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400204410.37/warc/CC-MAIN-20200922063158-20200922093158-00370.warc.gz | 337,648,677 | 7,646 | ## Conversion formula
The conversion factor from cubic inches to liters is 0.0163870640693, which means that 1 cubic inch is equal to 0.0163870640693 liters:
1 in3 = 0.0163870640693 L
To convert 35.5 cubic inches into liters we have to multiply 35.5 by the conversion factor in order to get the volume amount from cubic inches to liters. We can also form a simple proportion to calculate the result:
1 in3 → 0.0163870640693 L
35.5 in3 → V(L)
Solve the above proportion to obtain the volume V in liters:
V(L) = 35.5 in3 × 0.0163870640693 L
V(L) = 0.58174077446015 L
The final result is:
35.5 in3 → 0.58174077446015 L
We conclude that 35.5 cubic inches is equivalent to 0.58174077446015 liters:
35.5 cubic inches = 0.58174077446015 liters
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 liter is equal to 1.7189786996244 × 35.5 cubic inches.
Another way is saying that 35.5 cubic inches is equal to 1 ÷ 1.7189786996244 liters.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that thirty-five point five cubic inches is approximately zero point five eight two liters:
35.5 in3 ≅ 0.582 L
An alternative is also that one liter is approximately one point seven one nine times thirty-five point five cubic inches.
## Conversion table
### cubic inches to liters chart
For quick reference purposes, below is the conversion table you can use to convert from cubic inches to liters
cubic inches (in3) liters (L)
36.5 cubic inches 0.598 liters
37.5 cubic inches 0.615 liters
38.5 cubic inches 0.631 liters
39.5 cubic inches 0.647 liters
40.5 cubic inches 0.664 liters
41.5 cubic inches 0.68 liters
42.5 cubic inches 0.696 liters
43.5 cubic inches 0.713 liters
44.5 cubic inches 0.729 liters
45.5 cubic inches 0.746 liters | 531 | 1,875 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2020-40 | latest | en | 0.774201 |
https://community.cartalk.com/t/http-action-publicbroadcasting-net-cartalk-posts-list-1976010-page/47899 | 1,606,241,977,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141176922.14/warc/CC-MAIN-20201124170142-20201124200142-00717.warc.gz | 251,725,661 | 7,320 | 4/24/2010-
RE: TAKING CURVES RESPONSES > 12/08
by: jtsanders 12/06/2008 4:12:22 PM
Re: Re: Re: Re: TAKING CURVES
I agree. There is no centrifugal force, only centripital and your description is correct. Remember that forces are not scalar quantities, but directed (vector) quantities. If the positive dierction is towrds the origin of the arc, the negative direction is away from the origin.
by: texases 12/08/2008 1:15:46 PM
by: neburex 12/13/2008 3:35:53 PM
When you turn it takes more power to maintain the same speed. As you turn the wheel for the curve you’ve already slowed a little. Adding a little gas to maintain the same speed you had before the turn makes for a better balanced car and then use the steering only to make the small correction necessary to maintain the turn. Sorry that this so long, but the best I can do given that there are books written on just these themes.
MY FINAL WORD ON THE MATTER-(DAN CAMMAROTO)
WITH A FEW EXCEPTIONS -NOTED ABOVE- I JUST WANTED TO SAY THAT YOU GUYS ARE A BUNCH OF TURKEYS ! ! !
I ORIGINALLY THOUGHT THIS WAS GOING TO TOM & RAY BUT APPARENTLY IT WENT OUT TO THE GENERAL POPULATION OF “CAR GUYS”
WANNA-BE S.
SOME OF YOU WERE OBJECTIVE, SOME WERE NASTY. AND SOME -THOUGH FEW- WERE ASTUTE.
I SUGGEST THAT ALL YOU GUYS WHO PUT ME DOWN OR TRIED TO ‘HELP’ -EXCEPT FOR THOSE WHO BENIGNLY DISAGREED (THOUGH YOU TOO WERE INCORRECT)- SHOULD PROBABLY STOP
DRIVING ALTOGETHER.
WHILE I WILL NOT SAY THAT I DO THINGS
‘MY WAY’ OR NOT, I WILL SAY THAT -NOW
THAT SOME TIME HAS PASSED- I WAS, BY & LARGE, REALLY TICKED AT MOST OF YOU BOZOS.
PLEASE JUST STOP DRIVING -(LORD KNOWS WHAT OTHER INCORRECT THINGS YOU DO ON THE ROAD !!! )- AND, BY ALL MEANS, DON’T BOTHER TO RETALIATE.
I?M DONE W/ THIS SECTION OF THE SITE, SO YOUR NASTY ?BACK-TALK’, WILL
GO UNREAD & UNHEEDED. SINCERELY, DC
Hmmm…Despite Mr. Cammaroto’s claims to the contrary, I would be willing to make a small wager that he will return to read the responses to his…spirited…comments.
So, Mr. Cammaroto, what has transpired since your last post—more than one year ago–on this topic?
Why did it take you so long to return to this topic?
And, why have your position on the topic and your level of civility changed so drastically in the interim?
Back on December 13, 2008, you posted the following:
“12/14/08- Geez, Fellas, I feel like I been drawn-and-quartered AND
tarred-and-feathered. Mr. Cammaroto -me- drives a humble 2004 Dodge SX2 .0, FYI.
OK, Alright ! I get it. I’ll mend my ways (and ‘stay out of town’ !).
Just one question for all of you- My letter is originally addressed to
’CAR-CARE P-R-O-F-E-S-S-I-O-N-A-L-S’. None of you fine gents advised in
your admonitions whether or not you were / are ‘professionals’- That is:
technicians, engineers, auto repairmen.
Not that there’s anything wrong with that. . . .
It’s just that I was looking for individuals who actually work on cars
for a living or are perhaps driving instructors.
But, anyway, thanks again for your time, energy and concern.
Sincerely, Dan Cammaroto”
May I suggest that you cut back on your caffeine intake?
Okay, lets see if I can piece this together…
The OP posted a discussion a year ago on the question of whether centrifugal force exists. Pursuant to its relationship to cornering. We debated.
A year later the OP is lambasting some of the respondants for being rude. Unfortunately, the OP has not been kind enough to link us to the original thread.
He wanted to hear only from “professionals”…which he defines as either technicians, engineers and auto repairment or just guys who work on cars…I’m unsure which.
Okay, I’m lost. Perhaps a year from now we’ll get whatever the question is (was?) resolved. Unless, of course, they legalize marjuana and the OP’s nerves calm down.
Signed: A. Turkey
PostScript: nice job tracking this down, VDC.
“Okay, I’m lost. Perhaps a year from now we’ll get whatever the question is (was?) resolved. Unless, of course, they legalize marjuana and the OP’s nerves calm down.”
I’m lost, too, and I’m not sure that I’m smart enough to find the discussion from long ago. BTW, Thorazine has a calmng influence as well, but it needs to be prescribed by a psychiatrist.
You mean a doctor…it has to be prescribed by a doctor…or a pharmacist on marijuana. | 1,161 | 4,293 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-50 | latest | en | 0.800603 |
https://www.physicsforums.com/threads/a-b-c-d-a-x-b-x-c-x-d.222994/ | 1,575,857,387,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540517156.63/warc/CC-MAIN-20191209013904-20191209041904-00185.warc.gz | 828,648,011 | 17,388 | # A + B + C + D = A x B x C x D
## Main Question or Discussion Point
Are there any natural number solutions for
A + B + C + D = A x B x C x D
besides
{1, 1, 2, 4}?
What class of equation does that above belong to?
## Answers and Replies
Suppose there is another solution set A', B', C', D' with larger numbers.
A' = A + n (=1+n)
B' = B + m (=1+m)
C' = C + o (=2+o)
D' = D + p (=4+p)
where n,m,o,p are some nonnegative integers that you add to the original numbers to get the new solution numbers.
Then
A' x B' x C' x D' = (A + n)(B + m)(C + o)(D + p)
= ABCD + ABCp + ABoD + ABop + AmCD + AmCp + AmoD + Amop + nBCD + nBCp + nBoD + nBop + nmCD + nmCp + nmoD + nmop
If none of n,m,o, or p are zero, then you can just match terms to find that
ABCD + ABCp + ABoD + ABop + AmCD + AmCp + AmoD + Amop + nBCD + nBCp + nBoD + nBop + nmCD + nmCp + nmoD + nmop
> A + n + B + m + C + o + D + p
= A' + B' + C' + D'
So the multiplication is too big for another solution to work if all the additive terms are positive.
Now it gets a little tricky when some of the additive terms are terms (n,m,o,p) are zero. If all 4 are zero (n=m=o=p=0), then we just have the origional solution, so we can ignore that possibility.
If 3 are zero, then
A + B + C + D = A x B x C x D and
A' + B + C + D = A' x B x C x D so subtracting the two, we have
A - A' = (A - A') x B x C x D
which can only happen if B,C,D are all 1, so we can ignore that case too.
This leaves the 2 cases left: either 1 or 2 of the additive numbers (n,m,o,p) is zero. I will just choose particular variables to set to zero, but you could switch them around.
If n=0, but no others, then
A' x B' x C' x D'
= ABCD + ABCp + ABoD + ABop + AmCD + AmCp + AmoD + Amop
= (A + 0)(BCD) + (B + m)(ACp) + (C + o)(AmD) + (D + p)(ABo) + Amop
= A' x (BCD) + B' x (ACp) + C' x (AmD) + D' x (ABo) + Amop
> A' + B' + C' + D'
If n and m are both zero (n=m=0) but no others, then
A' x B' x C' x D'
= ABCD + ABCp + ABoD + ABop
= A + B + C + D + ABCp + ABoD + ABop, remember A+B+C+D = ABCD!
> A + B + C + o + D + p, by matching terms
= A' + B' + C' + D'
Soooo..... there are no larger solutions, and smaller possibilities ({1,1,1,1}, {1,1,1,2},...{1,1,2,3}) can be checked by hand.
Looks like the solution you got is the only one.
lol, this proof can be made much simpler:
A' x B' x C' x D' = (A + n)(B + m)(C + o)(D + p)
= ABCD + ABCp + ABoD + AmCD + nBCD + other nonnegative terms
= A + B + C + D + ABCp + ABoD + AmCD + nBCD + other (remember ABCD=A+B+C+D still)
> A + n + B + m + C + o + D + p
= A' + B' + C' + D'
Last edited:
Thanks, Nick. The simplest proofs are the best. Occam dices, slices, chops and minces.
Do you know what class of equation the original one was?
No clue. Perhaps someone else will know?
I can say that the above proof strategy will work for any number of variables in the same form (ABCDEFG... = A+B+C+D+E+F+G+...).
For n variables,
X1 x X2 x X3 x ... x Xn = X1 + X2 + X3 + ... + Xn
if n is even then {1, 1, 1, ..., 2, n-2} is the solution. eg:
1,1,2,4
1,1,1,1,2,6
1,1,1,1,1,1,2,8
why? Because
1 + 1 + ... + 2 + n = (n - 2) + 2 + n = 2*n
You could also do this by induction on n.
I'm not sure about odd numbers though.
Edit: duh, it works for odd numbers too.
Last edited:
morphism
Homework Helper
Do you know what class of equation the original one was?
It's an example of a (non-linear) Diophantine equation (in 4 variables).
Odds
For odd numbers of variables there seems to be more than one solution.
for
$$A + B + C + D + E = A * B * C * D * E$$
{1,1,2,2,2} works, as does
{1,1,1,3,3}
Ahh yes, interesting. Technically for the proof I posted, you need to check all combinations of sets involving only the numbers 1,2,3, and 4. I just (stupidly) assumed there were no other small solutions.
I wonder how many other solutions there are for big odd N, as for N variables the proof does not rule out solutions that are combinations of 1,2,3,...,N. | 1,432 | 3,931 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2019-51 | latest | en | 0.738454 |
https://www.fishlore.com/aquariumfishforum/threads/conversion-rate.6205/ | 1,580,299,772,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251796127.92/warc/CC-MAIN-20200129102701-20200129132701-00137.warc.gz | 862,732,431 | 20,634 | conversion rate
jj
New Member
When people talk in gallons for their tanks, what is the general concensus? One conversion table says mine is 10 US and another 12 Imperial. Just curious as two gallons is quite a bit different.
Isabella
Fishlore VIP
I'm not quite sure about what you're asking. A general consensus on what? In fish-keeping, we're generally concerned with Liters, US gallons, and UK gallons. According to the Fish Lore calculator ( https://www.fishlore.com/ConversionCalculator.htm ), 1 US gallon = almost 3.8 Liters, and 1 US gallon = 0.83 UK gallons.
jj
New Member
It may have seemed a 'dim' question, :-[ but as this is a global forum, if you advise on stocking per gallon, 2 gals is a bit if a difference and I notice that almost everyone talks in gallons. I was simply curious. I am however aware at the end of the day liters are the all important.
Isabella
Fishlore VIP
I am sorry for that. You must be from Europe (other than UK)? The reason most people here use US gallons is that a lot of people on Fish Lore are from the US. I understand it may be confusing to you if you're used to Liters. I would probably also be confused if I used a website where everyone talks only in Liters. But hey ... this is why Fish Lore has the Calculator So that we can all understand each other better
For example, if you hear somwhere that you should have a filter that pumps at least 10 times the volume of your tank per 1 hour, and if you tell me your tank is 110 Liters ... Well, you go to the Calculator and it tells you 110L = almost 30 gallons. But it doesn't matter in that case if it's Liters or Gallons, because 10 x the volume of a tank per hour is just that. In your case, you'd need a filter that pumps 1,100 Liters per hour, while in American gallons it would have to be a filter that pumps 300 gallons per hour.
I hope this helps you somewhat
jj
New Member
yep, I get it! Volume is volume no matter what calculator you use...it was just one of those things that I was curious about. I had to smile the other day when I saw someone buy a tank same as mine from same shop and was told it held 2gals more. That in turn lead me to the question. Thanks for replying.
No problemo | 540 | 2,205 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2020-05 | longest | en | 0.967575 |
https://math.stackexchange.com/questions/109136/finding-limits-algebraically?noredirect=1 | 1,624,282,986,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488273983.63/warc/CC-MAIN-20210621120456-20210621150456-00091.warc.gz | 353,796,030 | 41,287 | # Finding limits algebraically
I am going over my test and I am not sure what I was supposed to do for this question.
$$\lim_{x \to -\infty} \frac {\sqrt {x^2 + 3x}}{3-2x}$$
I attempted to do the algebra but my algebra skills are too weak to do it so I resorted to logic. I stated that the top will reduce to a positive x and the bottom will reduce to a positive $2x$ giving $\frac {x}{2x}$ which will then give me $1/2$ but this answer is wrong. Apparently the answer is correct but my reasoning is wrong.
I don't know why it is wrong but it is incredibly frustrating to me that I can get the correct answer with logical conclusions but still get the question on the test wrong. This is my second time taking calculus and at this rate I am going to fail again, no matter how hard I try I just can't get it right for some reason.
I guess this could be a broader question, but how do you take a math test? It is a mystery to me and I have absolutely no idea what a teacher expects on tests. They showed us this method in class, but on a test it is incorrect. To further complicate things I got the epsilon delta problem wrong because I didn't show the absolute value and all that stuff at the start even though I was able to show the correct epsilon.
• Don't give up! Also, the "this could be a broader question..." IS a broader question - too much so for this site, in fact. If you ever want to discuss such things, maybe we could do it in chat. Ping me. – The Chaz 2.0 Feb 14 '12 at 3:39
• You do have to be careful with your ad-hoc reasoning. For instance, $\lim_{x\rightarrow\infty} [\sqrt{x^2+x}-x]=\frac12$. You might be inclined to believe this limit "reduces" to $x-x=0$. – dls Feb 14 '12 at 4:10
Your intuition is actually correct. Most likely, your teacher wanted the 'proper' steps and justification for that intuition. To be fair to your teacher, intuition might not always be right, and without proper mathematical justification, what you write is just English :-)
We can rewrite as
$$\lim_{y \to \infty} \frac{\sqrt{y^2 - 3y}}{3 + 2y} =$$ $$\lim_{y \to \infty} \frac{y\sqrt{1 - \frac{3}{y}}}{y(\frac{3}{y} + 2)} =$$ $$\lim_{y \to \infty} \frac{\sqrt{1 - \frac{3}{y}}}{(\frac{3}{y} + 2)} = \frac{1}{2}$$
as $\frac{3}{y} \to 0$ as $y \to \infty$.
If you have an expression of the form where you have a polynomial raised to a fractional power divided by another polynomial raised to a fractional power, such as $$\sqrt{x^2-3x}\over 3-2x$$ and you're taking a limit at infinity or negative infinity: $$\tag{1} \lim_{x\rightarrow-\infty} {\sqrt{x^2-3x}\over 3-2x}.$$ Then identify the "highest power of $x$" that you see taking into account any outer fractional powers. I'll call this the dominant $x$ term of the expression. This will be the larger of the dominant terms of the numerator and denominator. If the dominant terms are the same in the numerator and denominator, they are the dominant term for the entire expression.
Back to our limit $(1)$:
In the denominator, the dominant $x$ term is $x$; in the numerator the dominant $x$ term is $\sqrt {x^2}=x$.
So, the dominant $x$ term of the entire expression is just $x$.
Now divide both top and bottom of the limit expression by $x$. Note that $x$ is negative since we are taking the limit at $-\infty$; so, we have to be careful when bringing $x$ into the radical:
$${\sqrt{x^2-3x}\over 3-2x} ={{\sqrt{x^2-3x}\over x}\over {3-2x\over x}} ={-{\sqrt {x^2-3x \over x^2 }}\over {{3\over x}-2}} ={-{\sqrt {1-{3\over x} }}\over {{3\over x}-2}}.$$ Then take the limit: $$\lim_{x\rightarrow-\infty} {\sqrt{x^2-3x}\over 3-2x} =\lim_{x\rightarrow-\infty}{-{\sqrt {1-{3\over x} }}\over {{3\over x}-2}}= {1\over 2}.$$
This is entirely rigorous and always works.
The process is particularly simple when taking a limit at infinity of a rational expression:
For example:
\eqalign{ \lim_{x\rightarrow\infty} {x^2-2x^7-1\over 3x^2+2x+3x^7} &=\lim_{x\rightarrow\infty} { {x^2\over x^7}-{2x^7\over x^7}-{1\over x^7} \over {3x^2\over x^7}+{2x\over x^7}+{3x^7\over x^7}}\cr &=\lim_{x\rightarrow\infty} { {1\over x^5}-{2 }-{1\over x^7} \over {3 \over x^5}+{2 \over x^6}+{3}}\cr &=-{2\over3}. }
It's instructive to explicitly point out that the method employed by Aryabhata (and elaborated in David's answer) is just a special case of computing limits by employing power series expansion, using Taylor/Laurent/Puiseux series. E.g the transformation $\sqrt{x^2-3x} = x \sqrt{1-3/x}$ to obtain the "dominant term" $x$ amounts to computing the leading term in the Laurent series at $x = +\infty$
$$(x^2 - 3x)^{1/2}\: =\ x (1-3/x)^{1/2}\: =\ x -\frac{3}2 - \frac{9}8 x^{-1} - \frac{27}{16} x^{-2} + \:\cdots$$
or, perhaps more familiar, in terms of $\:z = 1/x\:$ at $\:z = 0^{+}$
$$(z^{-2}-3z^{-1})^{1/2}\: =\ \left(\frac{1-3z}{z^2}\right)^{1/2} =\ \frac{1}z - \frac{3}2 -\frac{9}8 z - \frac{27}{16} z^2 +\:\cdots$$
Note that the limit of a Laurent series $f$ is simply the limit of its dominant leading term ${\rm L}(f)$
$$\lim_{z\:\to\: 0^{+}}\: (a z^n + b z^{n+1} + c z^{n+2}+\:\cdots\:)\ = \lim_{z\:\to\: 0^{+}} a z^n\: (1 + \frac{b}a z + \frac{c}a z^2 + \:\cdots)\ = \lim_{z\:\to\: 0^{+}} az^n$$
Thus for purposes of computing limits we need only compute the leading term of the series.
Further, since the leading term map is multiplicative ${\rm L}(f/g) = {\rm L}(f)/{\rm L}(g)$, to compute the limit of a quotient it suffices to compute the limit of the quotient of their leading terms. This explains the algebra behind the heuristics in the other answers. The conceptual background will become clearer when one studies (higher-rank) valuation theory. In fact, employing such ideas yields effective algorithms for computing limits of a large class of elementary functions. To learn more google "transseries" and see this answer. | 1,853 | 5,801 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2021-25 | latest | en | 0.960209 |
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[Caml-list] "Nasty" functions and memory usage
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Date: 2001-03-19 (15:27) From: Alex Baretta Subject: Re: [Caml-list] "Nasty" functions and memory usage
```Hendrik Tews wrote:
>
> let f0 = function x,y -> x,y;;
> let f1 = function x,y -> f1(f0 x, f0 y);;
> let ...
> let f5 = function x,y -> f5(f4 x, f4 y);;
>
> Can you post the correct program? For this I get
>
> # let f1 = function x,y -> f1(f0 x, f0 y);;
>
> Characters 26-28:
> Unbound value f1
>
> And if I insert let rec's f5 is typechecked in less than a
> second.
>
That's right. I screwed up.
let f0 = function x -> x,x;;
let f1 = function x,y -> f0 ( f0 x, f0 y);;
let f2 = function x,y -> f1 ( f1 x, f1 y);;
let f3 = function x,y -> f2 ( f2 x, f2 y);;
let f4 = function x,y -> f3 ( f3 x, f3 y);;
let f5 = function x,y -> f4 ( f4 x, f4 y);;
f0 doubles the size of the type of its argument. f1 doubles twice
the size (*4) of the type of its arguents. f2 multiplies by 4
twice (*16) the size of the type of its argument. f3 multiplies by
16 twice the size of the type of its argument (*256). f4
multiplies its arguments type size by 65536. f5 by 65536**2, which
is 4294967296, or 2**(2**5)). Basically, one sees immediately that
fn has space complexity = k*2**(2**n). A nice double exponential.
Find me a computer capable of f7, now... (3.4*10**38 times the
size of the argument)
Now, my question is, how come the memory usage of the Ocaml (3.00)
front end does not grow with time as it attempts to compute the
type of, say, f6?
Alex
-------------------
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``` | 610 | 1,900 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2018-26 | latest | en | 0.770552 |
http://slidegur.com/doc/59283/a-c | 1,503,366,694,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886109803.8/warc/CC-MAIN-20170822011838-20170822031838-00610.warc.gz | 382,035,953 | 15,525 | ### A c
```An Introduction to
Multivariate Models
Sarah Medland
SGDP Summer School
July 2010
•Univariate Model
Twin Model
•Bivariate Model
•Multivariate Models
Hypothesised
Sources of
Variation
Topics of Discussion:
Extensions to multiple variables
(3 or more)
Choosing between :
Model
Equations
Path
Diagrams
Matrix
Algebra
Path Tracing
Rules
Predicted Var/Cov
from Model
Cholesky Decomposition
Common Pathway model
Independent Pathway model
Observed Var/Cov
from Data
Structural Equation Modelling
(SEM)
Multivariate analysis
Univariate analysis: genetic and environmental
influences on the variance of one trait
Bivariate analysis: genetic and environmental
influences on the covariance between two traits
Multivariate analysis: genetic and environmental
basis of the covariance between multiple traits
Multiple phenotypes
Comorbid phenotypes
Diagnostic subtypes (e.g. anxiety: panic, social, separation)
Different dimensions (e.g. cognitive abilities)
Different raters
Self report, Mother-report, Teacher-report, Observational
Longitudinal data
Time-point 1, Time-point 2, Time-point 3, Time-point 4
Multivariate models
Specific
variance
Trait 1
Trait
2
Common
variance
Trait
3
Why do these phenotypes covary?
Multivariate models
Different models have different assumptions in the
nature of shared causes among multiple phenotypes
Cholesky Decomposition (Correlated factors solution)
Genetic and environmental factors on different variables
correlate
Independent pathway model
Specific and Common genetic and environmental causes
Common pathway model
Latent Psychometric factor mediates common genetic and
environmental effects
Cholesky Decomposition
Twin 1
A E
Mum
C
A E
Teacher
C
A E
Examiner
C
Twin 2
A E
A E
A E
A E
Child
Mum
Teacher
Examiner
C
C
C
Nvar Number of A, C and E Factors
C
A E
Child
C
The A Structure
Twin 1
A
Mum
A
Teacher
A
Examiner
Twin 2
A
Child
A
Mum
A
Teacher
A
Examiner
Number of A paths: [ nvar*(nvar+1) ] / 2
(4*5)/2 = 10
A
Child
The C Structure
Twin 1
Twin 2
Number of C paths: [ nvar*(nvar+1) ] / 2
(4*5)/2 = 10
Mum
Teacher
Examiner
Child
C
C
C
C
Mum
C
Teacher
Examiner
Child
C
C
C
The E Structure
Twin 1
E
Mum
E
Teacher
E
Examiner
Twin 2
E
Child
E
Mum
E
Teacher
E
Examiner
Number of E paths: [ nvar*(nvar+1) ] / 2
(4*5)/2 = 10
E
Child
Cholesky Decomposition
Twin 1
A E
A E
A E
Twin 2
rMZ = 1;rDZ = 0.5
A E
Mum
Teacher
Examiner
Child
C
C
C
C
A E
A E
A E
Mum
Teacher
Examiner
Child
C
C
C
C
rMZ/DZ = 1
A E
Correlated factors solution
am=am2
A
E
A
aT
Mum
A
C
C
am
E
Teacher
rA (M-T)
rA (M-E)
rA (M-C)
E
A
C
aE
Examiner
E
C
aC
Child
rA (T-E)
rA (T-C)
rA (E-C)
Correlated factors solution
cm=cm2
A
E
C
A
E
C
cm
Mum
A
E
C
rC (M-T)
rC (M-E)
rC (M-C)
E
C
cE
cT
Teacher
A
Examiner
cC
Child
rC (T-E)
rC (T-C)
rC (E-C)
Correlated factors solution
em=em2
A
E
C
A
E
C
em
Mum
A
E
A
C
rE (M-T)
rE (M-E)
rE (M-C)
C
eE
eT
Teacher
E
Examiner
eC
Child
rE (T-E)
rE (T-C)
rE (E-C)
Correlated factors solution
Assumptions
1. Each variable (e.g. Mother-rating) is influenced
by a set of genetic, shared and non-shared
environmental factors
2. The factors associated with each variable are
allowed to correlate with each other through rA, rC
and rE
3. Correlations among phenotypes are a function of
rA, rC and rE and the standardized A, C and E
paths connecting them
Independent & Common
Partition variance between variables:
1) Common variance:
variance that is shared by all measured variables
2) Specific variance:
variance that is not shared by the measured variables
S
Trait 1
C
Trait 2
S
C
C
C
Trait 3
S
What might common and specific variance represent?
1) Comorbid phenotypes: (e.g. anxiety subtypes)
Common variance = general liability to emotional reactivity
Specific variance = symptom-specific risks
2) Different raters: (e.g. mother, teacher, child reports)
Common variance = pervasive liability to reported behaviour
Specific variance = situation-specific behaviour
Independent pathway model
rMZ = 1 rDZ = 0.5
Twin 1
rMZ / DZ = 1
A
Teacher
Mum
C
E
Examiner
Child
C
A
C
A
E
Twin 2
E
Mum
C
Teacher
C
E
A
C
A
E
A
Examiner
Child
C
E
A
C
A
E
C
E
A
C
A
E
E
Independent pathway model
Assumptions
1. Each variable (e.g. mother rating) has variation
that is shared with other variables
“Common” genetic and environmental factors
2. Each variable is also influenced by unique
variance not shared with other variables
“Specific” genetic and environmental factors
3. Covariation among phenotypes may be due to
the same genetic or environmental causes
Independent pathway model
Example
To examine the etiology of comorbidity: e.g. the
separate symptom clusters of anxiety and
depression are influenced by the same genetic
factors
Conclusion: genes act largely in a non-specific
way to influence the overall level of psychiatric
symptoms. Separable anxiety and depression
symptom clusters in the general population are
largely the result of environmental factors
(Kendler KS et al., Arch Gen Psych, 1987)
Common pathway model
rMZ = 1 rDZ = 0.5
Twin 1
Twin 2
rMZ / DZ = 1
A
C
E
E
C
Latent
factor
fmum
fteacher
Teacher
Mum
Latent
factor
Examiner
A
A
E
fmum
fchild
fexaminer
Child
C
C
Mum
fteacher
Teacher
C
E
A
C
A
A
E
fchild
fexaminer
Examiner
Child
C
E
A
C
A
E
C
E
A
C
A
E
E
Common pathway model
Assumptions
1. Each variable (e.g. mother rating) has variation
that is shared with other variables and variation
that is specific
2. Common genetic and environmental variance is
captured by a latent psychometric factor (e.g.
pervasive or situation independent behaviour;
general liability to anxiety)
3. Covariation among phenotypes is due to the
effects of the common psychometric factor on
each variable
Common pathway model
Example
To study the etiology of Comorbidity: e.g.
experimentation and novelty seeking, used as
indices of a latent behavioral disinhibition trait
> h2 =0.84
Conclusion: a variety of adolescent problem
behaviours may share a common underlying
genetic risk
[Young et al., Am. J. Med. Genet. (Neuropsychiatric Genet.), 2000].
Observed Statistics and df
(Theoretical) Observed Summary Statistics
Maximum Likelihood Analysis using summary matrices as
input
Theoretical degrees of freedom
N observed summary statistics – N estimated parameters
(Actual) Observed Statistics
Full Maximum Likelihood Analysis using raw data as input
Number of available data points
(Actual) degrees of freedom
N observed statistics – N estimated parameters
Exercise 1
We will run these models on the 4 antisocial measures
1. How many observed summary statistics will there be?
a. Consider size of observed variance-covariance matrix
b. Consider number of observed means
Note: there are MZ and DZ pairs
2. How many parameters are estimated for
a. Cholesky Decomposition?
b. Independent pathway model?
c. Common pathway model?
Note: don’t forget means are estimated in each model too
3. How many theoretical degrees of freedom will each
model have?
Note: The variance of the common latent factor is constrained to be 1
Note: Df = observed statistics – estimated parameters
Variance - Covariance Matrix
8 x 8 matrix
T1V1
T1V2 T1V3
T1V4
T2V1
T2V2 T2V3 T2V4
T1V1
var
T1V2
cov
var
T1V3
cov
cov
var
T1V4
cov
cov
cov
var
T2V1
cov
cov
cov
cov
var
T2V2
cov
cov
cov
cov
cov
var
T2V3
cov
cov
cov
cov
cov
cov
var
T2V4
cov
cov
cov
cov
cov
cov
cov
var
Variance - Covariance Matrix
MZ Twins
8 x 8 symmetrical matrix
= (8x9)/2= 36 summary statistics
DZ Twins
8 x 8 symmetrical matrix
= (8x9)/2= 36 summary statistics
Total number = 72
Observed means
1 x 8 matrix
T1V1
T1V2
T1V3
T1V4
T2V1
T2V2
T2V3
T2V4
Mean1 Mean2 Mean3 Mean4 Mean1 Mean2 Mean3 Mean4
MZ Twins
1 x 8 Full matrix = 8 summary statistics
DZ Twins
1 x 8 Full matrix = 8 summary statistics
Total number = 16
Total number of Summary statistics
Summary statistics = variance-covariances + means
Summary statistics = 72 + 16=
88
Parameter estimates and DF
Model
Cholesky
Number of
estimated
parameters
??
Observed
Summary
statistics
88
Independent
pathway
88
Common
pathway
88
Theoretical
degrees of
freedom
Cholesky Decomposition
A E
A E
A E
A E
Mum
Teacher
Examiner
Child
C
C
C
C
A E
A E
A E
Mum
Teacher
Examiner
Child
C
C
C
C
(4x5)/2 =10 for A, C and E = 30
means (equated for birth-order
and) zygosity = 4
A E
Parameter estimates and DF
Model
Cholesky
Number of
estimated
parameters
34
Observed
summary
statistics
88
Independent ??
pathway
88
Common
pathway
88
Theoretical
degrees of
freedom
54
Independent pathway model
Twin 1
Twin 2
A
C
E
E
C
A
12
Teacher
Mum
Examiner
Child
Mum
Teacher
Examiner
Child
12
C
A
C
A
E
C
E
A
C
A
E
C
E
A
C
A
+4 means = 4
E
C
E
A
C
A
E
E
Parameter estimates and DF
Model
Observed
summary
statistics
88
Theoretical
degrees of
freedom
54
Independent 28
Pathway
88
48
Common
Pathway
88
Cholesky
Number of
estimated
parameters
34
??
Common pathway model
rMZ = 1 rDZ = 0.5
Twin 1
Twin 2
rMZ / DZ = 1
A
C
E
E
C
A
3
Latent
factor
fmum
fteacher
Teacher
Mum
C
A
C
A
E
Latent
factor
4Examiner
Child
A
C
A
Mum
E
E
C
A
+4 means = 16
fteacher
Teacher
fexaminer
Examiner
fchild
Child
NB. There is a
C
C
constraint – how is it
Aaccounted
E
A Mx ?
E
C for in
C
12
E
fmum
fchild
fexaminer
E
A
E
Parameter estimates and DF
Model
Observed
summary
statistics
88
Theoretical
degrees of
freedom
54
Independent 28
pathway
88
60
Common
pathway
89 (88+1)
66
Cholesky
Number of
estimated
parameters
34
23
Comparing models
Correlated
Factors
Independent
pathway
Common
pathway
→
Most restricted →
Most parsimonious →
Fewest parameters
openMx Script for
Independent Pathway model
Note: Scripts are at the end of these slides in your handout
Independent pathway model
rMZ = 1 rDZ = 0.5
Twin 1
rMZ / DZ = 1
A
Teacher
Mum
C
E
Examiner
Child
C
A
C
A
E
Twin 2
E
Mum
C
Teacher
C
E
A
C
A
E
A
Examiner
Child
C
E
A
C
A
E
C
E
A
C
A
E
E
Script
nvar <- 4
#number of variables
nf <- 1
#number of factors
ACE_Independent_Model <- mxModel("ACE_Independent",
mxModel("ACE",
mxMatrix( type="Full", nrow=nvar, ncol=nf, free=TRUE,
values=.6, name="ac" ),
mxMatrix( type="Full", nrow=nvar, ncol=nf, free=TRUE,
values=.6, name="cc" ),
mxMatrix( type="Full", nrow=nvar, ncol=nf, free=TRUE,
values=.6, name="ec" ),
rMZ = 1 rDZ = 0.5
rMZ / DZ = 1
Twin 1
A
C
acv1 acv2 acv3
V1
V2
E
Twin 2
E
C
acv4
V3
A
acv1 acv2 acv3 acv4
V4
V1
V2
V3
Matrix ac = path coefficient additive genetic
parameters of the common A factor
Ac
acv1
Variable 1
acv2
ac =
Variable 2 Full 4x1
acv3
Variable 3
acv4
Variable 4
V4
rMZ = 1 rDZ = 0.5
Twin 1
Twin 2
rMZ / DZ = 1
A
C
E
E
ccv1 ccv2 ccv3 ccv4
V1
V2
V3
C
A
ccv1 ccv2 ccv3 ccv4
V4
V1
V2
V3
Matrix cc = path coefficient shared environmental
parameters of the common C factor
Cc
ccv1
Variable 1
ccv2
cc =
Variable 2 Full 4x1
ccv3
Variable 3
ccv4
Variable 4
V4
rMZ = 1 rDZ = 0.5
Twin 1
rMZ / DZ = 1
A
C
E
E
ecv2
V2
V3
C
A
ecv4
ecv1
V1
Twin 2
ecv3 ecv4
V4
ecv1 ecv2 ecv3
V1
V2
V3
V4
Matrix ec = path coefficient non-shared environmental
parameters of the common E factor
Ec
ecv1
Variable 1
ecv2
ec =
Variable 2 Full 4x1
ecv3
Variable 3
ecv4
Variable 4
Script
mxMatrix( type="Diag", nrow=nvar, ncol=nvar, free=TRUE,
values=4, name="as" ),
mxMatrix( type="Diag", nrow=nvar, ncol=nvar, free=TRUE,
values=4, name="cs" ),
mxMatrix( type="Diag", nrow=nvar, ncol=nvar, free=TRUE,
values=5, name="es" ),
Matrix as= path coefficients genetic parameters of the
specific A factors
AV1 AV2 AV3 AV4
as =
V2
V1
asV1
asV2
Variable 1
0
asV2
0
0
asV3
Variable 2
Variable 3
0
0
0
V3
A
E
asV4
A
C
A
E
E
V4
asV4
C
A
C
A
V3
asV3
asV2
asV1
C
E
V2
V1
asV4
asV3
Diag 4x4
Variable 4
V4
C
C
A
asV1
E
Cross-twin cor between A and C factors omitted
C
E
A
C
A
E
E
Matrix cs= path coefficients shared environment
parameters of the specific C factors
CV1 CV2 CV3 CV4
cs =
V2
V1
csV1
Variable 1
0
csV2
0
0
csV3
Variable 2
Variable 3
0
0
0
V3
A
C
A
E
E
A
A
E
V3
V4
csV2
csV1
C
C
V2
V1
csV4
csV3
C
Variable 4
V4
csV2
csV1
csV4
Diag 4x4
E
A
csV3
C
A
C
E
csV4
E
C
A
C
A
E
E
Matrix es= path coefficients non-shared environment
parameters of the specific E factors
EV1 EV2 EV3 EV4
es =
V2
V1
esV1
A
C
A
C
E
esV1
Variable 1
0
esV2
0
0
esV3
Variable 2
Variable 3
0
0
0
V3
esV2
E
Variable 4
V4
esV3
C
A
C
A
esV4
E
Diag 4x4
V2
V1
esV4
esV1
E
A
C
A
E
C
V3
esV2
E
V4
esV3
A
C
A
C
E
esV4
E
Script
mxAlgebra(ac %*% t(ac) + as %*% t(as), name="A" ),
mxAlgebra(cc %*% t(cc) + cs %*% t(cs), name="C" ),
mxAlgebra(ec %*% t(ec) + es %*% t(es), name="E" ),
Matrix A= variance components of the common A factor
plus variance components of the specific A factors
ac %*% t(ac) + as %*% t(as)
Ac
acv1
acv2
ac =
acv3
acv4
Variable 1
Variable 2
Variable 3
Variable 4
Full 4x1
ac %*% t(ac)= 4x1 * 1x4
=
ac2V1
acV1acV2
acV1acV3
acV1acV4
acV2acV1
ac2V2
acV2acV3
acV2acV4
acV3acV1
acV3acV2
ac2V3
acV3acV4
acV4acV1
acV4acV2
acV4acV3
ac2V4
4x4
ac %*% t(ac) + as %*% t(as)
AV1 AV2 AV3 AV4
as =
asV1 0
0
0
0
asV2
0
0
0
0
asV3
0
0
0
0
Variable 1
Variable 2
Variable 3
asV4
Diag 4x4
Variable 4
as %*% t(as) = 4x4 * 4x4
as2V1 0
=
0
0
0
0
as2V2 0 0
0
0
0
as2V3
0
0
0
0
as2V4
4x4
Matrix A= variance components of the common A factor
plus variance components of the specific A factors
ac %*% t(ac) + as %*% t(as)
ac2V1+ as2V1 acV1acV2
A=
acV1acV3
acV1acV4
acV2acV1
ac2V2 + as2V2 acV2acV3
acV2acV4
acV3acV1
acV3acV2
ac2V3 + as2V3 acV3acV4
acV4acV1
acV4acV2
acV4acV3
ac2V4 + as2V4
4x4
Matrix A= variance components of the common A factor
plus variance components of the specific A factors
ac %*% t(ac) + as %*% t(as)
ac2V1+ as2V1 acV1acV2
A=
acV1acV3
acV1acV4
acV2acV1
ac2V2 + as2V2 acV2acV3
acV2acV4
acV3acV1
acV3acV2
ac2V3 + as2V3 acV3acV4
acV4acV1
acV4acV2
acV4acV3
ac2V4 + as2V4
4x4
Matrix C= variance components of the common C factor
plus variance components of the specific C factors
cc %*% t(cc) + cs %*% t(cs)
cc2V1+ cs2V1 ccV1ccV2
C=
ccV1ccV3
ccV1ccV4
ccV2ccV1
cc2V2 + cs2V2 ccV2ccV3
ccV2ccV4
ccV3ccV1
ccV3ccV2
cc2V3 + cs2V3 ccV3ccV4
ccV4ccV1
ccV4ccV2
ccV4ccV3
cc2V4 + cs2V4
4x4
Matrix E= variance components of the common E factor
plus variance components of the specific E factors
ec %*% t(ec) + es %*% t(es)
ec2V1+ es2V1 ecV1ecV2
E=
ecV1ecV3
ecV1ecV4
ecV2ecV1
ec2V2 + es2V2 ecV2ecV3
ecV2ecV4
ecV3ecV1
ecV3ecV2
ec2V3 + es2V3 ecV3ecV4
ecV4ecV1
ecV4ecV2
ecV4ecV3
ec2V4 + es2V4
4x4
Script
mxAlgebra(A+C+E, name="V" ),
mxMatrix( type="Iden", nrow=nvar, ncol=nvar,
name="I"),
mxAlgebra( solve(sqrt(I*V)), name="iSD"),
mxMatrix( type="Full", nrow=1, ncol=nvar,
free=TRUE, values= 0, name="M" ),
mxAlgebra( cbind(M,M), name="expMean"),
Matrix V =
Total variance/covariance matrix of
measured variables
V = A+C+E
V[1,1] = ac2V1+ as2V1+ cc2V1+ cs2V1+ ec2V1+ es2V1
V[2,1] = acV2asV1+ ccV2csV1+ ecV2esV1
V[3,1] = acV3asV1+ ccV3csV1+ ecV3esV1
...
iSD
mxMatrix( type="Iden", nrow=nvar, ncol=nvar,
name="I"),
mxAlgebra( solve(sqrt(I*V)), name="iSD"),
Multiply the var/cov matrix by an identity matrix
Yeilds a matrix with variances on the diagonals & zeros on the
off diagonals
The inverse of this yeilds a matrix with standard deviations on
the diagonals & zeros on the off diagonals
1
I=
0
var1 cov12
0
1
V=
cov12 var2
var1 0
I*V=
0
var2
sd1 0
iSD=
0
sd2
Script
mxAlgebra( rbind ( cbind(A+C+E , A+C),
cbind(A+C , A+C+E)),
name="expCovMZ" ),
mxAlgebra( rbind ( cbind(A+C+E , 0.5%x%A+C),
cbind(0.5%x%A+C , A+C+E)),
name="expCovDZ" )
),
Total Common variance/covariance:
A+C+E
ACEvarv1 ACEcov12 ACEcov13 ACEcov14
ACEcov21 ACEvarv2 ACEcov23 ACEcov24
ACEcov31 ACEcov32 ACEvarv3 ACEcov34
ACEcov41 ACEcov42 ACEcov43 ACEvarv4
Common covariance between MZ twins:
A+C
ACvarv1
ACcov12
ACcov13
ACcov14
ACcov21
ACvarv2
ACcov23
ACcov24
ACcov31
ACcov32
ACvarv3
ACcov34
ACcov41
ACcov42
ACcov43
ACvarv4
Common covariance between DZ twins:
.5A+C
.5ACvarv1 .5ACcov12 .5ACcov13 .5ACcov14
.5ACcov21 .5ACvarv2 .5ACcov23 .5ACcov24
.5ACcov31 .5ACcov32 .5ACvarv3 .5ACcov34
.5ACcov41 .5ACcov42 .5ACcov43 .5ACvarv4
A+C+E
A+C
T1V1
T1V2
T2V3
A+C
A+C+E
T1V4
T2V1
T2V2
T2V3
T2V4
T1V1
ACEvarV1 ACEcov ACEcov ACEcov
ACvarV1 ACcov ACcov
ACcov
T1V2
ACEcov ACEvarV2 ACEcov ACEcov
ACcov
ACcov
T1V3
ACEcov ACEcov ACEvarV3 ACEcov
ACcov ACcov ACvarV3 ACcov
T1V4
ACEcov ACEcov ACEcov ACEvarV4
ACcov ACcov ACcov ACvarV4
ACvarV2 ACcov
T2V1
ACvarV1 ACcov ACcov
ACcov
ACEvarV1 ACEcov ACEcov ACEcov
T2V2
ACcov
ACcov
ACEcov ACEvarV2 ACEcov ACEcov
T2V3
ACcov ACcov ACvarV3 ACcov
ACEcov ACEcov ACEvarV3 ACEcov
T2V4
ACcov ACcov ACcov ACvarV4
ACEcov ACEcov ACEcov ACEvarV4
ACvarV2 ACcov
A+C+E
.5A+C
T1V1
T1V2
T2V3
T1V4
.5A+C
A+C+E
T2V1
T2V2
T2V3
T2V4
T1V1
ACEvarV1 ACEcov ACEcov ACEcov.5ACvarV1 .5ACcov .5ACcov .5ACcov
T1V2
ACEcov ACEvarV2 ACEcov ACEcov.5ACcov .5ACvarV2 .5ACcov .5ACcov
T1V3
ACEcov ACEcov ACEvarV3 ACEcov.5ACcov .5ACcov .5ACvarV3 .5ACcov
T1V4
ACEcov ACEcov ACEcov ACEvarV4.5ACcov .5ACcov .5ACcov
.5ACvarV4
T2V1 .5ACvar .5ACcov .5ACcov .5ACcov ACEvar ACEcov ACEcov ACEcov
V1
V1
T2V2 .5ACcov .5ACvarV2 .5ACcov .5ACcov ACEcov ACEvarV2 ACEcov ACEcov
cov .5ACcov .5ACvar
cov ACEcov ACEcov ACEvar
cov
.5AC
.5AC
ACE
V3
V3
T2V3
cov .5ACcov .5ACcov
var ACEcov ACEcov ACEcov ACEvar
.5AC
.5AC
V4
V4
T2V4
Script
mxModel("MZ",
mxData( observed=mzData, type="raw" ),
mxFIMLObjective( covariance="ACE.expCovMZ",
means="ACE.expMean", dimnames=selVars )
),
Script
mxModel(“DZ",
mxData( observed=dzData, type="raw" ),
mxFIMLObjective( covariance="ACE.expCovDZ",
means="ACE.expMean", dimnames=selVars )
),
Mx Script for
Common pathway model
Note: Scripts are at the end of these slides in your handout
Common pathway model
rMZ = 1 rDZ = 0.5
Twin 1
Twin 2
rMZ / DZ = 1
A
C
E
E
C
Latent
factor
fmum
fteacher
Teacher
Mum
Latent
factor
Examiner
A
A
E
fmum
fchild
fexaminer
Child
C
C
Mum
fteacher
Teacher
C
E
A
C
A
A
E
fchild
fexaminer
Examiner
Child
C
E
A
C
A
E
C
E
A
C
A
E
E
Script
ACE_Common_Model <- mxModel("ACE_Common",
mxModel("ACE",
mxMatrix( type="Lower", nrow=nf, ncol=nf,
free=TRUE, values=.6, name="al" ),
mxMatrix( type="Lower", nrow=nf, ncol=nf,
free=TRUE, values=.6, name="cl" ),
mxMatrix( type="Lower", nrow=nf, ncol=nf,
free=TRUE, values=.6, name="el" ),
rMZ = 1 rDZ = 0.5
rMZ / DZ = 1
A
C
E
E
C
A
al
al
Latent
factor
1
Latent
factor
1
Matrix al=path coefficient for A parameter on common
latent factor
al =
A
al
Latent factor
Full 1x1
rMZ = 1 rDZ = 0.5
rMZ / DZ = 1
A
C
E
cl
1
el
Latent
factor
E
C
el
A
cl
Latent
factor
1
Matrix cl=path coefficient for C parameter on common
C
latent factor
cl =
cl
Common latent factor Full 1x1
Matrix el=path coefficient for E parameter on common
E
latent factor
el =
el
Common latent factor Full 1x1
Script
mxAlgebra( al %*% t(al) + cl %*% t(cl) + el %*% t(el),
name="CovarLP" ),
mxAlgebra( diag2vec(CovarLP), name="VarLP" ),
mxMatrix( type="Unit", nrow=nf, ncol=1, name="Unit"),
mxConstraint (VarLP == Unit),
CovarLP= variance of the latent factor
al %*% t(al) + cl %*% t(cl) + el %*% t(el)
A
al =
Common latent factor Full 1x1
al
X*X’ = 1x1 * 1x1
A=
a2l
1x1
CovarLP= variance of the latent factor
al %*% t(al) + cl %*% t(cl) + el %*% t(el)
CovarLP =
a2l+c2l+e2l
1x1
mxAlgebra( diag2vec(CovarLP), name="VarLP" )
VarLP =
a2l+c2l+e2l
1x1
mxConstraint (VarLP == Unit),
VarLP =
a2l+c2l+e2l
=1
Script
mxMatrix( type="Diag", nrow=nvar, ncol=nvar,
free=TRUE, values=.4, name="as" ),
mxMatrix( type="Diag", nrow=nvar, ncol=nvar,
free=TRUE, values=.4, name="cs" ),
mxMatrix( type="Diag", nrow=nvar, ncol=nvar,
free=TRUE, values=.5, name="es" ),
Matrix as= path coefficients genetic parameters of the
specific A factors
AV1 AV2 AV3 AV4
as =
V2
V1
asV1
asV2
Variable 1
0
asV2
0
0
asV3
Variable 2
Variable 3
0
0
0
V3
A
E
asV4
A
C
A
E
E
V4
asV4
C
A
C
A
V3
asV3
asV2
asV1
C
E
V2
V1
asV4
asV3
Diag 4x4
Variable 4
V4
C
C
A
asV1
E
Cross-twin cor between A and C factors omitted
C
E
A
C
A
E
E
Matrix cs= path coefficients shared environment
parameters of the specific C factors
CV1 CV2 CV3 CV4
cs =
V2
V1
csV1
Variable 1
0
csV2
0
0
csV3
Variable 2
Variable 3
0
0
0
V3
A
C
A
E
E
A
A
E
V3
V4
csV2
csV1
C
C
V2
V1
csV4
csV3
C
Variable 4
V4
csV2
csV1
csV4
Diag 4x4
E
A
csV3
C
A
C
E
csV4
E
C
A
C
A
E
E
Matrix es= path coefficients non-shared environment
parameters of the specific E factors
EV1 EV2 EV3 EV4
es =
V2
V1
esV1
A
C
A
C
E
esV1
Variable 1
0
esV2
0
0
esV3
Variable 2
Variable 3
0
0
0
V3
esV2
E
Variable 4
V4
esV3
C
A
C
A
esV4
E
Diag 4x4
V2
V1
esV4
esV1
E
A
C
A
E
C
V3
esV2
E
V4
esV3
A
C
A
C
E
esV4
E
Script
mxMatrix( type="Full", nrow=nvar, ncol=nf,
free=TRUE, values=1, name="f" ),
covariance due to the latent factor
Common Latent factor
f=
fv1
Variable 1
fv2
fv3
Variable 2
Variable 3
fv4
Variable 4
Full 4x1
Latent
factor
f v1
fv2
fv4
fv3
V2
V1
Latent
factor
V3
V4
C
A
C
A
E
fv1
fv2
V2
V1
A
C
A
E
V3
V4
C
C
E
fv4
fv3
E
A
C
A
E
C
E
A
C
A
E
E
Script
mxAlgebra( f %&% (al %*% t(al)) + as %*% t(as),
name="A" ),
mxAlgebra( f %&% (cl %*% t(cl)) + cs %*% t(cs),
name="C" ),
mxAlgebra( f %&% (el %*% t(el)) + es %*% t(es),
name="E" ),
Matrix A= variance components of the common A factor
plus variance components of the specific A factors
f %&% (al %*% t(al)) + as %*% t(as)
Common Latent factor
Full 4x1
fv1
f=
Latent A factor
al =
al
Full 1x1
fv2
fv3
fv4
fv1
=
fv2
fv3
fv4
*
al
*
al
* fv1 fv2 fv3 fv4
Matrix A= variance components of the common A factor
plus variance components of the specific A factors
f %&% (al %*% t(al)) + as %*% t(as)
fv1
=
fv2
fv3
*
al
*
al
* fv1 fv2 fv3 fv4
fv4
=
fv12al2
fv1fv2al2
fv1fv3al2 fv1fv4al2
fv2fv1al2
fv22al2
fv2fv3al2 fv2fv4al2
fv3fv1al2
fv3fv2al2
fv32al2
fv3fv4al2
fv4fv1al2
fv4fv2al2
fv4fv3al2
fv42al2
f %&% (al %*% t(al)) + as %*% t(as)
AV1 AV2 AV3 AV4
as =
asV1 0
0
0
0
asV2
0
0
0
0
asV3
0
0
0
0
Variable 1
Variable 2
Variable 3
asV4
Diag 4x4
Variable 4
as %*% t(as) = 4x4 * 4x4
as2V1 0
=
0
0
0
0
as2V2 0 0
0
0
0
as2V3
0
0
0
0
as2V4
4x4
Matrix A= variance components of the common A factor
plus variance components of the specific A factors
f %&% (al %*% t(al)) + as %*% t(as)
=
fv12al2+as12
fv1fv2al2
fv1fv3al2
fv1fv4al2
fv2fv1al2
fv22al2+as22
fv2fv3al2
fv2fv4al2
fv3fv1al2
fv3fv2al2
fv32al2+as32
fv3fv4al2
fv4fv1al2
fv4fv2al2
fv4fv3al2
fv42al2 +as42
Matrix C= variance components of the common C factor
plus variance components of the specific C factors
f %&% (cl %*% t(cl)) + cs %*% t(cs)
=
fv12cl2+cs12
fv1fv2cl2
fv1fv3cl2
fv1fv4cl2
fv2fv1cl2
fv22cl2+cs22
fv2fv3cl2
fv2fv4cl2
fv3fv1cl2
fv3fv2cl2
fv32cl2+cs32
fv3fv4cl2
fv4fv1cl2
fv4fv2cl2
fv4fv3cl2
fv42cl2 +cs42
Matrix E= variance components of the common E factor
plus variance components of the specific E factors
f %&% (el %*% t(el)) + es %*% t(es)
=
fv12el2+es12
fv1fv2el2
fv1fv3el2
fv1fv4el2
fv2fv1el2
fv22el2+es22
fv2fv3el2
fv2fv4el2
fv3fv1el2
fv3fv2el2
fv32el2+es32
fv3fv4el2
fv4fv1el2
fv4fv2el2
fv4fv3el2
fv42el2 +es42
Script
mxAlgebra( A+C+E, name="V" ),
mxMatrix( type="Iden", nrow=nvar, ncol=nvar,
name="I"),
mxAlgebra( solve(sqrt(I*V)), name="iSD"),
mxMatrix( type="Full", nrow=1, ncol=nvar, free=TRUE,
values= 0, name="M" ),
mxAlgebra( cbind(M,M), name="expMean"),
Script
mxAlgebra( rbind ( cbind(A+C+E , A+C),
cbind(A+C , A+C+E)),
name="expCovMZ" ),
mxAlgebra( rbind ( cbind(A+C+E , 0.5%x%A+C),
cbind(0.5%x%A+C , A+C+E)),
name="expCovDZ" )
),
Script
mxModel("MZ",
mxData( observed=mzData, type="raw" ),
mxFIMLObjective( covariance="ACE.expCovMZ",
means="ACE.expMean", dimnames=selVars )
),
Script
mxModel("DZ",
mxData( observed=dzData, type="raw" ),
mxFIMLObjective( covariance="ACE.expCovDZ",
means="ACE.expMean", dimnames=selVars )
),
Applications of multivariate genetic
modelling
Note: instructions and answer sheets for the practical
session are at the end of these slides in your handout
Sample characteristics
Aims of E-risk longitudinal study project:
To study etiology of Antisocial behaviour in 5-year-old twins (Erisk longitudinal study) using 4 independent sources of
information
Sample:
451 MZ twin pairs, 389 same-sex DZ twin pairs
Data:
Mothers, Teachers, Examiner-observer reports
(e.g. CBCL and DSM-IV items)
Child self-report (Berkeley Puppet Interview)
Twin 1 variables:
Twin 2 variables:
Mr1 Tr1 Er1 Sr1
Mr2 Tr2 Er2 Sr2
Research questions
Previous studies
Early childhood antisocial behaviour is a strong
prognostic indicator for poor adult mental
health. However, genetic ethiology is unknown
Current aims:
To examine the heritability of AB in children
using multiple informants to account for bias
Current analyses:
Test 3 multivariate models to examine the factor
structure common to the 4 measures of AB
1.
2.
3.
Within-twin cross-measure correlations
Twin correlations within-measure
Cross-twin cross-measure correlations
1.
2.
3.
4.
Run the saturated model (to obtain -2LL statistics
of multivariate models)
Test all three multivariate models, at the end of the
common pathway model run the model
comparison syntax and select the model of best-fit
(from comparison of sub-models)
Fill in the parameter estimates for each model in
path diagrams
Run confidence intervals for only the selected
elements (see handout)
Goodness-of-Fit measures
AIC (Akiake Information Criterion) = 2 - 2df
AIC is founded on ideas from information theory, gives a GoF
measure that penalises models for increasing complexity.
AIC can be used for nonnested models.
In comparing two models AIC = AICi – AICmin, where AICi is the
AIC value for model i, and AICmin is the AIC value of the ‘best’
model (Burnham and Anderson, 2002). As a rule of thumb:
∆AIC
Evidence
<2
3− 7
> 10
suggests substantial evidence for the model i
model i has considerably less support
model i is very unlikely
Goodness-of-Fit measures
BIC (Bayesian Information Criterion) = 2 - [df*ln(n)]
Takes sample size into account with increasingly negative values
corresponding to increasingly better fitting models.
BIC can be used to compare nonnested models.
In comparing two models the differences in BIC gives an
estimate of the strength of evidence in favour of the model with
the smaller BIC value (Raftery, 1995)
Grades of Evidence Corresponding to BIC difference between
Model 1 and Model 2:
BIC Difference
0−2
2−6
6 − 10
> 10
Evidence
Weak
Positive
Strong
Very Strong
Overall Goodness of Fit
Mx also produces a sample size adjusted BIC and Deviance
Information Criterion.
AIC, BIC and DIC are useful for testing non-nested models.
For example comparing a model with genetic dominance
(ADE) to one with shared environment effects (ACE).
Models with lower values are preferred.
Markon & Krueger carried out a simulation study that
indicated that BIC was preferable to AIC (especially with
large samples and complex models).
One hopes that the various fit statistics will converge to
select the same best fitting model.
References
Raftery, AE. (1995). Bayesian model selection in
social research. Sociological Methodology (ed. PV.
Marsden), Oxford, U.K.: Blackwells, pp. 111-196.
Burnham, KP & Anderson, DR. (2002). Model
Selection and Multimodel Inference: a practical
information-theoretic approach, 2nd edition.
Springer-Verlag, New York.,
Markon, KE & Kreuger,RF. (2004). Beh Gen, 34, pg
593 – 610.
Standardized Estimates
e.g. IP model
### Generate Multivariate Cholesky ACE Output ###
parameterSpecifications(ACE_Cholesky_Fit)
expectedMeansCovariances(ACE_Cholesky_Fit)
tableFitStatistics(ACE_Cholesky_Fit)
ACEpathMatrices <c("ACE.a","ACE.c","ACE.e","ACE.iSD","ACE.iSD %*%
ACE.a","ACE.iSD %*% ACE.c","ACE.iSD %*% ACE.e")
ACEpathLabels <c("pathEst_a","pathEst_c","pathEst_e","sd","stPathEst_a","stPat
hEst_c","stPathEst_e")
formatOutputMatrices(ACE_Cholesky_Fit,ACEpathMatrices,ACE
pathLabels,Vars,4)
References
Hewitt JK, Silberg JL, Neale MC, Eaves LJ, Erickson M. (1992).
The analysis of parental ratings of children’s behavior using
LISREL. Behavior Genetics, 22, 293-317.
van den Oord EJCG, Boomsma DI, & Verhulst FC (2000). A
study of genetic and environmental effects on the cooccurrence of problem behaviors in three-year-old twins.
Journal of Abnormal Psychology, 109, 360-372.
Arseneault, L., Moffitt, T.E., Caspi, A., Taylor, A., Rijsdijk, F.V.,
Jaffee, S., Ablow, J.C., & Measelle, J.R. (2003). Strong genetic
effects on cross-situational antisocial behavior among 5-yearold children according to mothers, teachers, examinerobservers, and twins’ self-reports. Journal of Child
Psychology and Psychiatry, 44, 832-848.
``` | 11,314 | 27,963 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2017-34 | longest | en | 0.795115 |
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• application and methods
sampling (statistics)
...information from every member of the population, as in biological or chemical analysis, industrial quality control, or social surveys. The basic sampling design is simple random sampling, based on probability theory. In this form of random sampling, every element of the population being sampled has an equal probability of being selected. In a random sample of a class of 50 students, for...
• opinion polling
public opinion: Probability sampling
Once the universe has been defined, a sample of the universe must be chosen. The most reliable method of probability sampling, known as random sampling, requires that each member of the universe have an equal chance of being selected. This could be accomplished by assigning a number to each person in the universe or writing each person’s name on a slip of paper, placing all the numbered or...
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statistics: Sample survey methods
...used to estimate it. For example, the difference between a population mean and a sample mean is sampling error. Sampling error occurs because a portion, and not the entire population, is surveyed. Probability sampling methods, where the probability of each unit appearing in the sample is known, enable statisticians to make probability statements about the size of the sampling error....
• sampling theory
sample preparation: Theory
...used to select the sample, most often falls into this category, which is called nonprobabilistic sampling. Such methods can never satisfactorily represent highly heterogeneous material. In contrast, probabilistic sampling methods are techniques in which all constituents of the material have some probability of being included. However, it is only in a correctly designed sampling plan that...
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KiwiC Memory Hazard Example
KiwiC: Memory Hazard Example
Here we show the default method for KiwiC to overcome structural hazards: by splitting conflicted pause blocks into microstates.
More-advanced implementations would schedule resources over multiple blocks to find an optimum schedule.
C# Source Code
```using System;
using System.Text;
using KiwiSystem;
public static class test4
{
static int limit = 10;
[Kiwi.OutputBitPort("dout")] static bool dout;
[Kiwi.InputBitPort("din")] static bool din;
static public void arraypart()
{
int odd = 0, even = 0; // These are V_0 and V_1 in the ast.cil file.
int[] arr = new int [] {0, 1, 222221, 5, 7, 8, 1121, 2021, 2048}; // arr=V_2
arr[2] = 2;
// V_3 is the current element of the array
// V_4 is a copy of V_2
int pr = 0;
foreach (int i in arr) // i=V_6
{
if (i%2 == 0)
even++;
else
odd++;
Kiwi.Pause();
Console.Write("{1} i={0}: ", i, pr++);
Console.WriteLine("so far {0} Odd Numbers, and {1} Even Numbers.", odd, even);
}
Console.WriteLine("Found {0} Odd Numbers, and {1} Even Numbers.", odd, even) ;
Kiwi.Pause();
}
public static void Main()
{
int j = 0;
arraypart();
if (false) while(true)
{
j = j + 1;
if (j >= limit) j = 0;
dout = din & (j % 2 > 0);
Console.WriteLine(" J={0}.", j);
Kiwi.Pause();
}
}
}
```
Hardware Generated (with structural hazards still present)
```module DUT(reset, clk, din, dout);
input reset;
input clk;
input din;
output dout;
reg signed [31:0] DRSX32SS_AX_CC_SOL_QC_SOL[8:0];
integer mpc10;
integer Ttar0_3_V_6;
integer Ttar0_3_V_4;
integer Ttar0_3_V_3;
integer Ttar0_3_V_1;
integer Ttar0_3_V_0;
always @(posedge clk ) begin
//Start HPR test4.exe
if (reset) begin
Ttar0_3_V_0 <= 32'd0;
Ttar0_3_V_3 <= 32'd0;
Ttar0_3_V_6 <= 32'd0;
Ttar0_3_V_4 <= 32'd0;
Ttar0_3_V_1 <= 32'd0;
mpc10 <= 32'd0;
end
else begin
case (mpc10)
2/*2:mpc10XS:":pcIS44:AG"*/: begin
if ((Ttar0_3_V_6<8)) if (!(!(DRSX32SS_AX_CC_SOL_QC_SOL[1+Ttar0_3_V_6]%2))) begin
mpc10 <= 4/*4:mpc10XS:":pcIS63:AG"*/;
Ttar0_3_V_0 <= 1+Ttar0_3_V_0;
Ttar0_3_V_4 <= DRSX32SS_AX_CC_SOL_QC_SOL[1+Ttar0_3_V_6];
end
else begin
Ttar0_3_V_1 <= 1+Ttar0_3_V_1;
Ttar0_3_V_4 <= DRSX32SS_AX_CC_SOL_QC_SOL[1+Ttar0_3_V_6];
end
else mpc10 <= 8/*8:mpc10XS:":pcIS56:AG"*/;
Ttar0_3_V_6 <= 1+Ttar0_3_V_6;
Ttar0_3_V_3 <= 1+Ttar0_3_V_3;
\$write("%d i=%d: ", Ttar0_3_V_3, Ttar0_3_V_4);
\$display("so far %d Odd Numbers, and %d Even Numbers.", Ttar0_3_V_0, Ttar0_3_V_1);
if ((Ttar0_3_V_6>=8))
\$display("Found %d Odd Numbers, and %d Even Numbers.", Ttar0_3_V_0, Ttar0_3_V_1);
end
4/*4:mpc10XS:":pcIS63:AG"*/: begin
if ((Ttar0_3_V_6<8)) if (!(!(DRSX32SS_AX_CC_SOL_QC_SOL[1+Ttar0_3_V_6]%2))) begin
Ttar0_3_V_0 <= 1+Ttar0_3_V_0;
end
else begin
mpc10 <= 2/*2:mpc10XS:":pcIS44:AG"*/;
Ttar0_3_V_1 <= 1+Ttar0_3_V_1;
end
else mpc10 <= 8/*8:mpc10XS:":pcIS56:AG"*/;
Ttar0_3_V_4 <= DRSX32SS_AX_CC_SOL_QC_SOL[1+Ttar0_3_V_6];
Ttar0_3_V_6 <= 1+Ttar0_3_V_6;
Ttar0_3_V_3 <= 1+Ttar0_3_V_3;
\$write("%d i=%d: ", Ttar0_3_V_3, Ttar0_3_V_4);
\$display("so far %d Odd Numbers, and %d Even Numbers.", Ttar0_3_V_0, Ttar0_3_V_1);
if ((Ttar0_3_V_6>8))
\$display("Found %d Odd Numbers, and %d Even Numbers.", Ttar0_3_V_0, Ttar0_3_V_1);
end
8/*8:mpc10XS:":pcIS56:AG"*/: \$finish(0);
endcase
if ((0==mpc10)) begin
mpc10 <= 2/*2:mpc10XS:":pcIS44:AG"*/;
Ttar0_3_V_1 <= 1;
Ttar0_3_V_4 <= 0;
Ttar0_3_V_6 <= 0;
Ttar0_3_V_3 <= 0;
Ttar0_3_V_0 <= 0;
DRSX32SS_AX_CC_SOL_QC_SOL[2] <= 2;
DRSX32SS_AX_CC_SOL_QC_SOL[7] <= 2021;
DRSX32SS_AX_CC_SOL_QC_SOL[5] <= 8;
DRSX32SS_AX_CC_SOL_QC_SOL[3] <= 5;
DRSX32SS_AX_CC_SOL_QC_SOL[1] <= 1;
DRSX32SS_AX_CC_SOL_QC_SOL[0] <= 0;
DRSX32SS_AX_CC_SOL_QC_SOL[4] <= 7;
DRSX32SS_AX_CC_SOL_QC_SOL[6] <= 1121;
DRSX32SS_AX_CC_SOL_QC_SOL[8] <= 2048;
end
if ((mpc10==8/*8:mpc10XS:":pcIS56:AG"*/)) mpc10 <= 1/*1:mpc10XS:":pcIS-10:AG"*/;
end
//End HPR test4.exe
end
// 9 array locations of width 32
// 192 bits in scalar variables
// Total state bits in module = 480 bits.
endmodule
// eof (HPR/LS Verilog)
```
Simulation Output without Restructuring
The following output is generated either by running the test4.exe on mono or Windows or by simulating the DUT.v component using an RTL simulator (e.g. Modelsim):
```0 i=0: so far 0 Odd Numbers, and 1 Even Numbers.
1 i=1: so far 1 Odd Numbers, and 1 Even Numbers.
2 i=2: so far 1 Odd Numbers, and 2 Even Numbers.
3 i=5: so far 2 Odd Numbers, and 2 Even Numbers.
4 i=7: so far 3 Odd Numbers, and 2 Even Numbers.
5 i=8: so far 3 Odd Numbers, and 3 Even Numbers.
6 i=1121: so far 4 Odd Numbers, and 3 Even Numbers.
7 i=2021: so far 5 Odd Numbers, and 3 Even Numbers.
8 i=2048: so far 5 Odd Numbers, and 4 Even Numbers.
Found 5 Odd Numbers, and 4 Even Numbers.
```
```module SIMSYS();
reg clk, reset;
initial begin reset = 1; clk = 0; # 400 reset = 0; end
always #100 clk = !clk;
DUT dut(.clk(clk), .reset(reset));
initial # 1000000 \$finish();
endmodule
// eof
```
Restructured to use an SRAM
After restructuring, a specific, single-ported RAM component is instantiated in the RTL with busses to access it. This can be included in the main always block (as shown below) or be a separate RTL module with a structural instantiation:
``` //Start HPR HPR_RAM_1_32
always @(posedge clk ) begin
end
//End HPR HPR_RAM_1_32
```
The generated RTL now looks like this DUT-R.v.
These busses are driven by the main code. The address bus is registered to suite pipelined RAM blocks typically used in FPGA and ASIC. The address needs to be one cycle earlier than the data (but does not appear as such in the timing diagram since AD0 has a blocking assign at the start of the always block).
As can be seen, additional clock cycles are used to sequence the operations on the memory port. Such splitting of a paus block into microstates would be a compile-time error in 'HardPauseMode' where the compiler must strictly obey the Kiwi.Pause calls embedded in the C# code.
Modified timing diagram: additional cycles of additional program counter (mpc10zz) sequence the writes:
31st May 2011. UP. | 2,323 | 6,088 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-30 | longest | en | 0.396481 |
http://articlesurfing.org/education/homeschooling_while_you_shop.html | 1,490,456,617,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218188962.14/warc/CC-MAIN-20170322212948-00342-ip-10-233-31-227.ec2.internal.warc.gz | 24,196,862 | 5,092 | ### Homeschooling While You Shop?
Yes, It can be done! We are all busy juggling multiple tasks at once, and it does not get any easier when you are homeschooling. Here is an idea to get your children to help out with a chore and provide them with a learning opportunity at the same time.
Before going to the grocery store, ask your children to help you with a shopping list. You can go around the house - the fridge, pantry, and the bathroom cabinet, etc.- and categorise the list. You can also go through the flyer and see what's on sale and check on the price differences from one store to the another. (You can also discuss distribution channel and marketing for older kids.) Bring a calculator (if your kids prefer doing math that way) for your shopping trip.
Once you are at the grocery store, let the children do the shopping while you supervise. Show them how to select items based on the quality and/or price. While you are comparing the price, do a quick math lesson, or if you buy multiple of one item, what the total price differences would be. In the produce section, discuss where fruits and vegetables are from, and why you find thing from that particular climate. You can also talk about environment and organic produce. Have children weight vegitables and ask them how much a pound and a half of grapes would be.
In the meat and seafood section, discuss where they come from, how they keep them fresh, and what would happen if they are not kept cold. Many seafood items are imported, so you may discuss geography. If you are cooking a roast that day, you may have the children use the meat themometer and determine how cooked it is, and if it is safe to eat.
While you are in line at the check out counter, take out your coupons and ask them how much you can save if you use the double coupons. If you buy 3 Klenex tissue boxes on sale for \$3, and have \$.75 off coupon, how much would each box costs? How about if you get "buy one get one free" can of soup for \$2.50 and have \$.50 off coupon?
Don't forget to recycle those soda cans and talk about aluminum, recycling, and environment! On the way home, talk about how much gas you used for what distance, and how you can save money and environment by reducing the number of trips you take each month. Why is the gas so expensive? Where does the oil come from? How about the new hybrid cars? If the kids are done talking, you can listen to a CD and complete you day with a music lesson!
Submitted by:
### Ura Kondo Rinaldi
Ura Kondo Rinaldi, M.A. is a mother of two and a contributing author for http://www.Zeroprep.com, a fast growing resource for homeschoolers with artticles, revews, tips and fun ideas to help make your daily planning easier. She is also a business consultant for BBI, http://www.bbiworld.com and advises clients to develop their business with creative and unique approaches.
This article may be freely distributed, provided that it is not edited or altered in any way. All links must remain active and this notice must also be included. Copyright 2006 Ura Kondo Rinaldi, M.A. - contact ura@zeroprep.com for more information.
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http://www.articlesurfing.org/education/homeschooling_while_you_shop.html | 810 | 3,578 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2017-13 | latest | en | 0.958561 |
http://fnaarccuneo.it/lqiu/course-3-chapter-4-functions-answer-key.html | 1,590,706,189,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347400101.39/warc/CC-MAIN-20200528201823-20200528231823-00598.warc.gz | 49,350,661 | 17,991 | Answers to Odd-Numbered Exercises25 Chapter 5. 5 Study Guide - Nervous system - KEY - Page 2 of 4 4. Ensure that students understand that the two \$50 amounts and the one \$11,000 amount skew the mode and mean, respectively. 4 Controlling Application Cursor Keys mode. 24 Find the inverse of the fu ion. CONTINUITY27 5. These materials include worksheets, extensions, and assessment options. There are an equal amount of raisins in each cup. Bring whatever supplies (loose leaf paper, notebook, pen, pencil, etc) you personally like to use to take notes. Use the table below to find videos, mobile apps, worksheets and lessons that supplement Glencoe Math Course 3 Volume 1 Common Core. 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Angles • Section 1. (a) (b) 39. Graph the function. mRNA - messenger RNA rRNA - ribosomal RNA tRNA - transfer RNA the complementary strand to DNA. -4-4-2 3 4 0 xy 3-2 1 0-xy 4 3 1-1-1 1 1 2 1 3 2 D = {-1, 1, 2, 3}; R = {-1, 1, 2} D = {-4, -2. These materials include worksheets, extensions, and assessment options. Test, Form 2A (continued) y O x-2-4-3. Graph paper File. 5 Direct Variation 4. Chapter 7 Resource Masters The Fast FileChapter Resource system allows you to conveniently file the resources you use most often. Chapter 39. 4 The Slope of a Line 4. 0 MB: Offline ZIP. There is also another standard sequence data. x whale to travel 30 kilometers. RECIPES A recipe calls for 2 !3 cups of milk. I do not know if there are questions about Swine Flu (H1N1 Virus) on the NCLEX but rest assured, since it has been such a big issue around the world, it will be on the NCLEX soon. 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It parallels the timing of the Mid-Chapter Quiz in the Student Edition and includes both multiple-choice and free-response questions. Close the CH2 Personal Budget file before moving on to 2. • Decide whether you Agree (A) or Disagree (D) with the statement. 6 (Factoring) covers basic factoring techniques and how to solve equations using those techniques along with the Zero Product Property of Real Numbers. Algebra 2 129 Chapter 5 Worked-out Solution Key Copyright © McDougal Littell Inc. Prophetic Dreams - Why and When The Missing Step. M 2 A The E-Commerce Difference. 50 per pound. 07x Evaluate each function for the given value. y 53 ?2x 4. 1234 in word form is one thousand two hundred thirty-four ten-thousandths. Created Date: 3/15/2011 10:45:50 PM. 2 Changing the action of the Home and End keys. Answers For Workbooks The answers for Chapter 4 of these workbooks can be found in the The Chapter 4 Resource Masters includes the core materials needed for Chapter 4. Answer: TRUE Chapter: 3. Tuples and Sequences ¶ We saw that lists and strings have many common properties, such as indexing and slicing operations. Answers (Anticipation Guide and Lesson 2-1) STEP 1 Chapter 2 3 Read the Lesson 1. YES! Now is the time to redefine your true self using Slader’s free Saxon Math Course 3 answers. I do not know if there are questions about Swine Flu (H1N1 Virus) on the NCLEX but rest assured, since it has been such a big issue around the world, it will be on the NCLEX soon. Solve Exercises 20 and 21 drawing a diagram. Graph the function =y 5 - −1 4 x to determine how many of each type of furniture. com, order ISBN 0-13-318549-4. Lesson 4: Powers of Monomials. 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Nov 30, 2010, 8. 0) Lesson Tutorials. Functions8 4. 2 Practice B for use with pages 125-129 1. CCNA 3 Chapter 4 Exam Answers 2017 - 2018 - 2019 100% Updated Full Questions latest. Course Descriptor Units Math Help Section 6. 44:37; 0; 7 years ago; Unit 3 test answer key. Write a function to represent the situation. For Exercises 6 and 7, find each function value. A function is a special type of relation in which, for every value of the. x 5 22, y 5 35 2. Sample answer: To divide the numerator of the complex fraction by the denominator, multiply the numerator by the reciprocal of the denominator. Our introduction to the R environment did not mention statistics, yet many people use R as a statistics system. 5 Packet Tracer – Implementing Basic Connectivity: 2. The AP Calculus Exam is on Tuesday, May 5, 2020, B-Day. Find the ratio of snails to the total number of animals. The term biology is derived from the Greek word βίος, vios, "life" and the suffix -λογία, -logia, "study of. 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Carries the "recipe" to the ribosome forms the structure of the ribosome transports amino acids to the ribosome. y = 5−x 6 + 1 6. I suggest you take notes on any questions you get incorrect. You can skip questions if you would like and come back to. Free math lessons and math homework help from basic math to algebra, geometry and beyond. Write the ordered pair that names point F. Chapter 5 Answers (continued) 42 Answers Algebra 2Chapter 5 63. loge 2000 32. read: 47 size: 1. How many deli 8 sandwiches did he make? Solve each inequality. This means that we already know how to graph functions. Skip to Main Content. Page 1 of 2 Contents vii CHAPTER2 CHAPTER STUDY GUIDE 66 2. 4) Answer Key. If the answer is correct, you will be taken to the next question. 5 million or 4. Contents CHAPTER 3 Human Body Systems Chapter Concept Map. Agents of Socialization. Motion Problems and Average vs. 2 Solving linear equations; 4. Students are to record inch 3 feet. 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Unit 4: Analyze and Graph Linear Equations, Functions and Relations Learning Objectives Lesson 1: Graphing Linear Equations Topic 1: Rate of Change and Slope Learning Objectives • Calculate the rate of change or slope of a linear function given information as sets of ordered pairs, a table, or a graph. Encoding is the act of getting information into our memory system through automatic or effortful processing. | 10,251 | 40,834 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2020-24 | latest | en | 0.891397 |
https://masomomsingi.com/charts/ | 1,713,347,698,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817146.37/warc/CC-MAIN-20240417075330-20240417105330-00797.warc.gz | 340,568,965 | 14,156 | # CHARTS
BAR CHARTS
These are some of the most popular ways of presenting statistics A bar chart consists of strips of different lengths each representing a numerical figure. The longer the bar the larger the figure it denotes
How to construct a bar chart
1. Decide on the scale in which the figures are to be represented example 1 cm for \$500.
2. Draw the axis – the lines which the scale is marked mark out the scale at regular intervals.
3. Decide on the order of the bars. If they show statistics over a period of time they should be arranged in chronological order.
4. Label the bars if necessary making sure that all writing is horizontal so that it can be read easily.
5. Give the chart a clear heading if possible at the top.
6. Give the source of the data used to reassure the reader that the figures are accurate.
PIE CHARTS
They are used to show the relative size of separate components of a whole. For example they can be used to show how the income of a country is spent.
GRAPHS
Graphs are a valuable form of visual communication they are a means of presenting data on the relationship between two constantly changing elements in the form of single line , the shape of which reveals the nature of the relationship at a glance.
HOW TO DRAW A GRAPH
1. Use a graph paper
2. choose the variables for each axis
3. Choose the scales for the two axis which will fit the graph paper
4. Mark off the axis at regular intervals
5. Plot the points of the graph.
6. join the points to draw the line graph
7. make sure each axis is labeled to show what information it gives and the scale used
8. Give the graph a clear title
(Visited 113 times, 1 visits today) | 374 | 1,667 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2024-18 | latest | en | 0.917262 |
https://www.urbanpro.com/cbse-class-3-maths-division-worksheets | 1,628,223,962,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152112.54/warc/CC-MAIN-20210806020121-20210806050121-00098.warc.gz | 1,103,385,792 | 11,929 | # Free CBSE Class 3 Maths Division Worksheets
Download free printable Division Worksheets to practice. With thousands of questions available, you can generate as many Division Worksheets as you want.
## Sample CBSE Class 3 Maths Division Worksheet Questions
1.
282 books are distributed equally to 7 classes ,then each class got ___ books and _____ left.
2.
Annual salary of Mukesh is Rs 14400. What is his monthly salary?
3.
How many apples are left if 26 apples are distributed equally into 6 students?
4.
How many years are there in 84 months?
5.
What number should be added to 87 such that it divisible by 15?
6.
What number should be added to 234 such that it divisible by 34?
7.
What number should be added to 515 such that it divisible by 65?
8.
What number should be subtracted from 139 such that it divisible by 17?
9.
What number should be subtracted from 67 such that it divisible by 19?
10.
Solve: 5874 รท 66 =
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# Problem 239. Project Euler: Problem 5, Smallest multiple
Solution 3403618
Submitted on 26 Oct 2020 at 20:07
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Fail
x = 20; y_correct = 232792560; assert(isequal(euler005(x),y_correct))
For x = 20 minimum divisible by all is 232792560
Output argument "y" (and maybe others) not assigned during call to "euler005". Error in Test1 (line 3) assert(isequal(euler005(x),y_correct))
2 Fail
x = 10; y_correct = 2520; assert(isequal(euler005(x),y_correct))
For x = 10 minimum divisible by all is 2520
Output argument "y" (and maybe others) not assigned during call to "euler005". Error in Test2 (line 3) assert(isequal(euler005(x),y_correct))
3 Fail
x = 12; y_correct = 27720; assert(isequal(euler005(x),y_correct))
For x = 12 minimum divisible by all is 27720
Output argument "y" (and maybe others) not assigned during call to "euler005". Error in Test3 (line 3) assert(isequal(euler005(x),y_correct))
4 Fail
x = 14; y_correct = 360360; assert(isequal(euler005(x),y_correct))
For x = 14 minimum divisible by all is 360360
Output argument "y" (and maybe others) not assigned during call to "euler005". Error in Test4 (line 3) assert(isequal(euler005(x),y_correct))
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Start Hunting! | 431 | 1,471 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2020-50 | latest | en | 0.742778 |
http://www.chegg.com/homework-help/questions-and-answers/cannonball-shot-bridge-30-meters-high-velocity-15-m-s-ball-land-draw-x-y-coordinates-two-c-q949665 | 1,474,826,500,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738660338.16/warc/CC-MAIN-20160924173740-00271-ip-10-143-35-109.ec2.internal.warc.gz | 368,666,292 | 14,134 | If a cannonball is shot off of a bridge that is 30 meters high and
will have a velocity of 15 m/s. Where will the ball land? Draw this
with x and y coordinates.
A) Two concrete blocks are dropped from a height of 20 ft. One
block has a mass of 10 g and the second weighs 25g. Without taking
into consideration wind resistances, which block will hit the ground
B) Draw a velocity-vs.-time graph for the following information. Use
the graph to find the acceleration from 0-3s, 3-4s, and 5-6s.
t=0s--- v= 0 m/s
t=1s--- v= 2 m/s
t=2s--- v= 4 m/s
t=3s--- v= 6 m/s
t=4s--- v= 6 m/s
t=5s--- v= 4 m/s
An outfielder wants to improve his catching skills and understand
motion. He is curious if he will have more time to move catching a pop
fly, or a ball hit at waist level. While at bat, he wants to know if a
high hit will allow him to run more or less bases, how does this
compare to hitting a ground ball? Provide an explanation to the player
and include motion diagrams. What can you do to help him understand
the concepts of 2-d motion?
A person jumps off a cliff at a speed of 2m/s. If this person lands
5 m away from the cliff, how high was the cliff? | 342 | 1,152 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2016-40 | latest | en | 0.944093 |
http://metamath.tirix.org/mpeascii/elkgen | 1,718,519,341,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861643.92/warc/CC-MAIN-20240616043719-20240616073719-00367.warc.gz | 21,444,819 | 1,929 | # Metamath Proof Explorer
## Theorem elkgen
Description: Value of the compact generator. (Contributed by Mario Carneiro, 20-Mar-2015)
Ref Expression
Assertion elkgen
|- ( J e. ( TopOn ` X ) -> ( A e. ( kGen ` J ) <-> ( A C_ X /\ A. k e. ~P X ( ( J |`t k ) e. Comp -> ( A i^i k ) e. ( J |`t k ) ) ) ) )
### Proof
Step Hyp Ref Expression
1 kgenval
|- ( J e. ( TopOn ` X ) -> ( kGen ` J ) = { x e. ~P X | A. k e. ~P X ( ( J |`t k ) e. Comp -> ( x i^i k ) e. ( J |`t k ) ) } )
2 1 eleq2d
|- ( J e. ( TopOn ` X ) -> ( A e. ( kGen ` J ) <-> A e. { x e. ~P X | A. k e. ~P X ( ( J |`t k ) e. Comp -> ( x i^i k ) e. ( J |`t k ) ) } ) )
3 ineq1
|- ( x = A -> ( x i^i k ) = ( A i^i k ) )
4 3 eleq1d
|- ( x = A -> ( ( x i^i k ) e. ( J |`t k ) <-> ( A i^i k ) e. ( J |`t k ) ) )
5 4 imbi2d
|- ( x = A -> ( ( ( J |`t k ) e. Comp -> ( x i^i k ) e. ( J |`t k ) ) <-> ( ( J |`t k ) e. Comp -> ( A i^i k ) e. ( J |`t k ) ) ) )
6 5 ralbidv
|- ( x = A -> ( A. k e. ~P X ( ( J |`t k ) e. Comp -> ( x i^i k ) e. ( J |`t k ) ) <-> A. k e. ~P X ( ( J |`t k ) e. Comp -> ( A i^i k ) e. ( J |`t k ) ) ) )
7 6 elrab
|- ( A e. { x e. ~P X | A. k e. ~P X ( ( J |`t k ) e. Comp -> ( x i^i k ) e. ( J |`t k ) ) } <-> ( A e. ~P X /\ A. k e. ~P X ( ( J |`t k ) e. Comp -> ( A i^i k ) e. ( J |`t k ) ) ) )
8 toponmax
|- ( J e. ( TopOn ` X ) -> X e. J )
9 elpw2g
|- ( X e. J -> ( A e. ~P X <-> A C_ X ) )
10 8 9 syl
|- ( J e. ( TopOn ` X ) -> ( A e. ~P X <-> A C_ X ) )
11 10 anbi1d
|- ( J e. ( TopOn ` X ) -> ( ( A e. ~P X /\ A. k e. ~P X ( ( J |`t k ) e. Comp -> ( A i^i k ) e. ( J |`t k ) ) ) <-> ( A C_ X /\ A. k e. ~P X ( ( J |`t k ) e. Comp -> ( A i^i k ) e. ( J |`t k ) ) ) ) )
12 7 11 syl5bb
|- ( J e. ( TopOn ` X ) -> ( A e. { x e. ~P X | A. k e. ~P X ( ( J |`t k ) e. Comp -> ( x i^i k ) e. ( J |`t k ) ) } <-> ( A C_ X /\ A. k e. ~P X ( ( J |`t k ) e. Comp -> ( A i^i k ) e. ( J |`t k ) ) ) ) )
13 2 12 bitrd
|- ( J e. ( TopOn ` X ) -> ( A e. ( kGen ` J ) <-> ( A C_ X /\ A. k e. ~P X ( ( J |`t k ) e. Comp -> ( A i^i k ) e. ( J |`t k ) ) ) ) ) | 1,031 | 2,038 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-26 | latest | en | 0.269775 |
https://stackoverflow.com/questions/72576712/remove-the-intersection-between-two-curves/72577360 | 1,660,167,702,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571210.98/warc/CC-MAIN-20220810191850-20220810221850-00781.warc.gz | 497,085,025 | 65,272 | # Remove the intersection between two curves
I'm having a curve (parabol) from 0 to 1 on both axes as follows:
I generate another curve by moving the original curve along the x-axis and combine both to get the following graph:
How can I remove the intersected section to have only the double bottoms pattern like this:
The code I use for the graph:
``````import numpy as np
import matplotlib.pyplot as plt
def get_parabol(start=-1, end=1, steps=100, normalized=True):
x = np.linspace(start, end, steps)
y = x**2
if normalized:
x = np.array(x)
x = (x - x.min())/(x.max() - x.min())
y = np.array(y)
y = (y - y.min())/(y.max() - y.min())
return x, y
def curve_after(x, y, x_ratio=1/3, y_ratio=1/2, normalized=False):
x = x*x_ratio + x.max() - x[0]*x_ratio
y = y*y_ratio + y.max() - y.max()*y_ratio
if normalized:
x = np.array(x)
x = (x - x.min())/(x.max() - x.min())
y = np.array(y)
y = (y - y.min())/(y.max() - y.min())
return x, y
def concat_arrays(*arr, axis=0, normalized=True):
arr = np.concatenate([*arr], axis=axis).tolist()
if normalized:
arr = np.array(arr)
arr = (arr - arr.min())/(arr.max() - arr.min())
return arr
x, y = get_parabol()
new_x, new_y = curve_after(x, y, x_ratio=1, y_ratio=1, normalized=False)
xx = concat_arrays(x, new_x, normalized=True)
yy = concat_arrays(y, new_y, normalized=True)
# plt.plot(x, y, '-')
plt.plot(xx, yy, '--')
``````
I'm doing a research on pattern analysis that requires me to generate patterns with mathematical functions.
Could you show me a way to achieve this? Thank you!
First off, I would have two different parabola functions such that:
``````import numpy as np
import matplotlib.pyplot as plt
x = np.linspace(-1, 1, 100)
y1 = np.add(x, 0.3)**2 # Parabola centered at -0.3
y2 = np.add(x, -0.3)**2 # Parabola centered at 0.3
``````
You can choose your own offsets for y1 and y2 depending on your needs.
And then it's simply take the min of the two arrays
``````y_final = np.minimum(y1, y2)
plt.plot(x, y_final, '--')
``````
• Appreciate your sugesstion but the end result is lacking the right side. Is there anyway to make it look exactly like a round "W"? Jun 10 at 17:10
• Yes, for that you need to modify the offsets so they are symmetric. Try: ``` y1 = np.add(x, 0.5)**2 # Parabola centered at -0.5 ; and y2 = np.add(x, -0.5)**2 # Parabola centered at 0.5 ``` Jun 10 at 17:23
• To have them closer together: y1 = np.add(x, -0.3)**2, and y2 = np.add(x, 0.3)**2 Jun 10 at 17:26
• Your second comment solved my problem! Can you update your answer then I could mark this as accepted? Jun 10 at 17:29
This involves curve fitting. You need to find the intersection part before you drop the values. Since the values of x and y have been normalized, we would have to determine exactly where the two datasets meet. We can see that they meet when x[i] >x[i+1]. Using your cobined `xx` and `yy` from the data provided, We therefore can do the following:
``````data_intersect = int(np.where(np.r_[0,np.diff(xx)] < 0)[0])
x1 = xx[:data_intersect]
x2 = xx[data_intersect:]
y1 = yy[:data_intersect]
y2 = yy[data_intersect:]
difference = np.polyfit(x1, y1, 2) - np.polyfit(x2,y2,2)
meet = np.roots(difference) # all points where the two curves meet
meet = meet[(meet < max(x1)) & (meet >min(x1))] # only point curve meet
xxx = np.r_[x1[x1<meet], x2[x2>meet]]
yyy = np.r_[y1[x1<meet], y2[x2>meet]]
plt.plot(xxx, yyy, '--')
``````
• Thank you this solution is what I have been looking for Jun 12 at 18:44 | 1,064 | 3,465 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2022-33 | latest | en | 0.64371 |
https://www.reasoningtricks.com/water-images/water-images-letters-numbers/ | 1,722,853,981,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640436802.18/warc/CC-MAIN-20240805083255-20240805113255-00812.warc.gz | 736,884,467 | 12,505 | # Water Images of Mixed Letter and Numbers (Non Verbal)
## Water Images of Mixed Letter and Numbers
In every Competitive exam, one of the most important section is Reasoning. If you get good score in Reasoning test then it will help you to achieve good marks in competitive exams. This can be achieve only if you have a very good reasoning skills. Basically reasoning is divided into few sections according to logic and type. One of them is Water Images of Mixed Letter and Numbers.
One of the most important chapter of Reasoning aptitude test is Water Images of Mixed Letter and Numbers. In your reasoning exam, few questions will definitely come from this topic. Here in this page we will discuss Water Images of Mixed Letter and Numbers reasoning with solution which will help you to solve Water Images of Mixed Letter and Numbers questions very easily and quickly. Reasoning shortcut tricks are nothing but to solve reasoning questions very fast and accurately. Here we will show you the method of “How to solve Reasoning using shortcut tricks”.
### Few Important things to Remember
Time is the most important factor in every competitive exams. You should complete your exam within the time frame. But in every competitive exam they also test your calculation ability within a given time frame. They tests, how fast a student can complete a question paper. For this reason, many students won’t complete their paper within the given time space. But for faster solution you need to use tricks of reasoning and Water Images of Mixed Letter and Numbers which will help you to solve government, bank and other exam papers quickly.
Here in www.ReasoningTricks.com we provide many reasoning shortcut tricks and online tests. Carefully learn those shortcut tricks. Go through every chapter of reasoning and learn. Then collect reasoning books form market or friends or collect PDF from online. You can also download reasoning eBooks from online if it’s free. Then solve the practice sets of those books and try to solve those reasoning MCQ questions using tricks which you have learn in www.ReasoningTricks.com. You will surely notice the difference between usage of reasoning tricks and non usage of reasoning tricks. Go through our reasoning notes on Water Images of Mixed Letter and Numbers will help to get success in your examination.
Now we will discuss all the details of Water Images of Mixed Letter and Numbers and important reasoning question and answer.
### Example #1
Choose the alternative which is closely resembles the water-image of the given combination.
UP4X9T
1.U
P
4X
9
T
2.U
P
4X9
T
3.
UP4X9T
4.U
P4X9T
3.
UP4X9T
### Example #2
Choose the alternative which is closely resembles the water-image of the given combination.
QS7AF5Z2
1.QS7
AF
5Z2
2.
QS7AF5Z2
3.
Q
S7AF
5
Z2
4.QS
7
A
F
5
Z
2
2.
QS7AF5Z2
### Example #3
Choose the alternative which is closely resembles the water-image of the given combination.
DZ6F74
1.D
Z
6F7
4
2.
D
Z6F
7
4
3.DZ
6F
7
4
4.
DZ6F74
4.
DZ6F74
### Example #4
Choose the alternative which is closely resembles the water-image of the given combination.
VAYU8346
1.
V
A
Y
U834
6
2.
V
AYU8
3
46
3.
VA
Y
U
834
6
4.
VAYU8346
4.
VAYU8346
### Example #5
Choose the alternative which is closely resembles the water-image of the given combination.
BIHAR32
1.BIHA
R
3
2
2.
HIB
A
R
3
2
3.
BIHAR32
4.
B
IHAR
3
2
3.
BIHAR32
### Example #6
Choose the alternative which is closely resembles the water-image of the given combination.
EN76DM
1.EN76
D
M
2.
E
N
76
D
M
3.EN
76
DM
4.
EN76DM
4.
EN76DM
### Example #7
Choose the alternative which is closely resembles the water-image of the given combination.
COZ821
1.128ZO
C
2.128
COZ
3.COZ8
21
4.
COZ821
4.
COZ821
### Example #8
Choose the alternative which is closely resembles the water-image of the given combination.
YTX9S56
1.
YTX9S56
2.YTX
9
S56
3.
Y
TX
9S56
4.YTX9S
56
1.
YTX9S56
### Example #9
Choose the alternative which is closely resembles the water-image of the given combination.
US91Q4M5W1
1.
US91Q4M5W1
2.
US
9
1
Q
4M5W
1
3.
U
S
91Q4M5W
1
4.
US
9
1Q4M5W1
1.
US91Q4M5W1
### Example #10
Choose the alternative which is closely resembles the water-image of the given combination.
5D0B6V3
1.
5
D
OB
6V
3
2.
5
DOB9V
3
3.
5
D
OB9
V
3
4.
5D0B6V3
4.
5D0B6V3
### Other types of Water Images
So, this is the basic idea about Water Images of Mixed Letter and Numbers. Now, if you have anything in your mind then feel free to ask us. Comment your thoughts in the comment section below. And also you can ask us any question regarding this topic, our team member will respond to your question as soon as possible.
Meanwhile, you can also like and follow our facebook page to get new updates. And, if you have any questions in your mind regarding this topic then please do comment in the comment section below. You can also send us messages on facebook. Our team will answer all your questions. | 1,356 | 4,937 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-33 | latest | en | 0.939071 |
http://www.pearltrees.com/violetfemmes/verbatim/id26211635 | 1,701,367,880,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100229.44/warc/CC-MAIN-20231130161920-20231130191920-00380.warc.gz | 75,982,567 | 22,255 | Verbatim
Fall - Idioms by The Free Dictionary. Characters to Words Conversion Calculator. Astronomy Answers: Coincidence Periods. Loading [MathJax]/jax/output/CommonHTML/fonts/TeX/fontdata.js 1.
Introduction At least 2500 years ago astronomers in Babylon discovered that solareclipses and lunar eclipses often recur under similar circumstances after 223 synodical months (counted according to the phases of the Moon, not according to the western calendar) or about 6585⅓ days. This period is nowadays called the saros. We call a series of eclipses that each occur one saros after the previous one a saros series. To get a solar eclipse or lunar eclipse, two independents series of periodic phenomena must come together, namely the proper phase of the Moon (New Moon for a solar eclipse, or Full Moon for a lunar eclipse) and passage of the Moon through a node of the orbit of the Moon around the Earth. 2.
We study the coming together of two different periodic phenomena A and B. PAPB=γ′=ab with a and b whole numbers. If an A and a B happen at the same time then we get the combination phenomenon Z. S≡ak+s0(modb) k=l(s−s0)+nb 3. C=PAδ. 16 Government Types - Infographic Facts. 16 Government Types.
Manninghamappendicesglossary. The words project. word list.the words project. SO MUCH TO TELL YOU: the 100 club. Study Guides and Strategies. Random words. The Dead Words - Weekly Fave on All My Faves. The Philosopher's Meme. Dictionary of Interjections (aww, oh, ah, eek, oops) Godwin's law. Mike Godwin (2010) Godwin's law (or Godwin's Rule of Nazi Analogies)[1][2] is an Internet adage asserting that "As an online discussion grows longer, the probability of a comparison involving Nazis or Hitler approaches 1" [2][3]— that is, if an online discussion (regardless of topic or scope) goes on long enough, sooner or later someone will compare someone or something to Hitler or Nazism.
Promulgated by American attorney and author Mike Godwin in 1990,[2] Godwin's Law originally referred, specifically, to Usenet newsgroup discussions.[4] It is now applied to any threaded online discussion, such as Internet forums, chat rooms and blog comment threads, as well as to speeches, articles and other rhetoric.[5][6] In 2012, "Godwin's Law" became an entry in the third edition of the Oxford English Dictionary.[7] Corollaries and usage History Godwin has stated that he introduced Godwin's law in 1990 as an experiment in memetics.[2] See also Notes
Poe's law. Confusion of parody and sincere expression Poe's law is an adage of Internet culture stating that, without a clear indicator of the author's intent, it is impossible to create a parody of extreme views so obviously exaggerated that it cannot be mistaken by some readers for a sincere expression of the views being parodied.[1][2][3] The original statement, by Nathan Poe, read:[1] Without a winking smiley or other blatant display of humor, it is utterly impossible to parody a Creationist in such a way that someone won't mistake for the genuine article.
Origin Poe's law is based on a comment written by Nathan Poe in 2005 on christianforums.com, an Internet forum on Christianity. The post was made during a debate on creationism, where a previous poster had remarked to another user: "Good thing you included the winky. Types of Reasoning. Disciplines > Argument > Types of Reasoning Reasoning within an argument gives the rationale behind why one choice, for example should be selected over another.
Types of reasoning include: Abduction: the process of creating explanatory hypotheses. Backwards Reasoning: Start from what you want and work back. Butterfly Logic: How people often argue. Note that these are not all mutually exclusive methods and several give different lenses onto overlapping areas. Learning. 100 Most beautiful words in the English language* 100 Beautiful and Ugly Words. Justenglish. Amazing — incredible, unbelievable, improbable, fabulous, wonderful, fantastic, astonishing, astounding, extraordinary.
Quotability
Verbata. Viewables. 12 Nouns With Absurdly Specific Definitions. Allusory. Skillsets. Wordsworth. WORD! Writing. | 948 | 4,123 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2023-50 | latest | en | 0.83005 |
https://www.scribd.com/document/22660620/GMAT-Diagnostic-Test-GMAT-Club-v2-8 | 1,563,927,318,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195530246.91/warc/CC-MAIN-20190723235815-20190724021815-00336.warc.gz | 820,010,589 | 61,739 | You are on page 1of 12
# Contributing to each other’s learning
## GMAT Club Diagnostic Test
This diagnostic test was put together by dedicated members of the GMAT Club. This is
our contribution to the next generation of GMAT Test takers – and in return we only ask
for your feedback - let us know how we did and what we can do better. Visit Diagnostic
a suggestion to improve this test.
How to use:
Diagnostic test consists of 45 Questions split into 15 sections. Each section consists of 3
questions arranged in an increasing difficulty order: 600-level question, 700-level and
750-level. You have an option to take the full test or you can just take the 600 and 700
level questions and save the 750-level questions. To accurately assess your results, make
sure you are spending no more than 2 minutes per question on average. You have 90
minutes to complete the test.
After you are done – please take a minute to share your feedback and scores here. Please
include how long it took you to complete the test and which questions you found of
interest or problems with.
## Thank you and Good Luck on the GMAT,
BB – Founder of GMATCLUB
-1-
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GMAT Club Diagnostic Test
Contributing to each other’s learning
## You have 90 minutes to complete the test.
ARITHMETIC
ROOTS 4. Is X a prime integer?
(1) |X| = 2
1. 324 + 289 = ? (2) X2 = 4
(A). 32 A. Statement (1) ALONE is
(B). 33 sufficient, but Statement (2)
(C). 34 ALONE is not sufficient
(D). 35 B. Statement (2) ALONE is
(E). 36 sufficient, but Statement (1)
ALONE is not sufficient
2. 36 + 64 + 5 2 + 20 = ? C. BOTH statements TOGETHER
are sufficient, but NEITHER
statement ALONE is sufficient
A. 19 + 20
D. EACH statement ALONE is
B. 19 20 sufficient
C. 145 E. Statements (1) and (2)
TOGETHER are NOT sufficient
D. 5 100 + 20
E. 7 5
## 3. If x is an integer and x × x − x = a , 5. Which of the following expressions has
which of the following must be true? the greatest value?
I. a is Even
II. a is Positive A. 99912
III. a is an Integer B. 10 30
C. 77710
A. I only
B. II only D. (− 20)24
C. III only E. ( 15 )40
D. I and II
E. None of the above
POWERS
-2-
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GMAT Club Diagnostic Test
Contributing to each other’s learning
## 6. What is the value of E. 22
2
1 1
×4
× 4
9. Is x greater than 1?
8 16 ?
1
1 1
> −1
2
−4
×2 (1)
x
64
1 1
(2) 5 > 3
A. 16 x x
B. 4
C. 2 A. Statement (1) ALONE is
1 sufficient, but Statement (2)
D. ALONE is not sufficient
2
B. Statement (2) ALONE is
1
E. sufficient, but Statement (1)
16 ALONE is not sufficient
C. BOTH statements TOGETHER
are sufficient, but NEITHER
statement ALONE is sufficient
NUMBER PROPERTIES D. EACH statement ALONE is
sufficient
7. Which of the following numbers is the E. Statements (1) and (2)
greatest? TOGETHER are NOT sufficient
## 1876452 SPECIAL CHARACTERS
A.
1876455
1883446 10. If S is the sum of the digits of a given
B.
1883449 number, T is the sum of the digit of S,
1883453 and G is the sum of digits in T. For
C. example S of 987 is 9+8+7 = 24, T of S
1883456
1883456 is 2+4 = 6 and G of 6 is 6. Therefore G of
D. 987 is 6. Which of the following has the
1883459 greatest G)?
1883491
E.
1883494 A. 94123
B. 91964
8. If m and n are integers, what is the C. 64678
smallest possible value of integer m D. 62355
m E. 45689
if = 0.36363636... ?
n
11. If N = 1234@ and @ represents the units
A. 3 digit, is N a multiple of 5?
B. 4
C. 7 (1) @! is not divisible by 5
D. 13 (2) @ is divisible by 9
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C. |x| - 2 > 2
A. Statement (1) ALONE is D. |2 + x| > 3
sufficient, but Statement (2) E. |x – 2| < 3
ALONE is not sufficient
B. Statement (2) ALONE is
sufficient, but Statement (1) 14. Is K a positive number?
ALONE is not sufficient
C. BOTH statements TOGETHER (1) |K3| + 1 > K
are sufficient, but NEITHER (2) K + 1 > |K3|
statement ALONE is sufficient
D. EACH statement ALONE is A. Statement (1) ALONE is
sufficient sufficient, but Statement (2)
E. Statements (1) and (2) ALONE is not sufficient
TOGETHER are NOT B. Statement (2) ALONE is
sufficient sufficient, but Statement (1)
ALONE is not sufficient
12. If A Ω B = A+B if A > B, and A Ω B = B C. BOTH statements TOGETHER
- A if A < B, then which of the are sufficient, but NEITHER
followings is the highest for statement ALONE is sufficient
1 1 1 1 D. EACH statement ALONE is
Ω Ω Ω ? sufficient
x y y x E. Statements (1) and (2)
TOGETHER are NOT sufficient
1 1
A. X = and Y =
15. If ( p !) = p ! , which of the following
2 3 p
1 1
B. X = and Y = could be the value(s) of p?
3 4
1 1 A. -1
C. X = and Y =
4 5 B. 0
1 1 C. 1
D. X = and Y =
5 4 D. -1 and 1
1 1 E. -1, 0, and 1
E. X = and Y =
4 2
STATISTICS
MODULES
16. Set T contains more than one element. Is
13. Which of the following inequalities must the median of set T greater than its mean?
be true if the values of x are between -1
and 5? (1) Set S has positive range.
(2) The elements of the set are not
A. |3 – x| < -3 consecutive integers
B. -1< |x| < 5
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## A. Statement (1) ALONE is A. Statement (1) ALONE is
sufficient, but Statement (2) sufficient, but Statement (2)
ALONE is not sufficient ALONE is not sufficient
B. Statement (2) ALONE is B. Statement (2) ALONE is
sufficient, but Statement (1) sufficient, but Statement (1)
ALONE is not sufficient ALONE is not sufficient
C. BOTH statements TOGETHER C. BOTH statements TOGETHER
are sufficient, but NEITHER are sufficient, but NEITHER
statement ALONE is sufficient statement ALONE is sufficient
D. EACH statement ALONE is D. EACH statement ALONE is
sufficient sufficient
E. Statements (1) and (2) E. Statements (1) and (2)
TOGETHER are NOT sufficient TOGETHER are NOT sufficient
## 17. Set S consists of N elements. If N>2,
what is the standard deviation of S? WORD PROBLEMS
(1) The mean and median of the set are
equal
(2) The difference between any two MIN/MAX
elements of the set is equal
19. Ten suits, each priced at a whole number
A. Statement (1) ALONE is of dollars, have an average price of \$350.
sufficient, but Statement (2) If the five cheapest suits are priced at
ALONE is not sufficient \$200 or less with at least one of them
B. Statement (2) ALONE is costing \$200, what is the maximum
sufficient, but Statement (1) possible price of the most expensive suit?
ALONE is not sufficient
C. BOTH statements TOGETHER A. \$890
are sufficient, but NEITHER B. \$1250
statement ALONE is sufficient C. \$1475
D. EACH statement ALONE is D. \$1694
sufficient E. \$2492
E. Statements (1) and (2)
TOGETHER are NOT sufficient 20. Three people each took 5 tests. If the
ranges of their scores in the 5 practice
18. Is the mean of set S greater than its tests were 17, 28 and 35, what is the
median? minimum possible range in scores of the
three test-takes?
(1) All members of S are consecutive
multiples of 3 A. 17
(2) The sum of all members of S equals B. 28
75 C. 35
D. 45
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## E. 80 24. Among 60 members of a club, 6p players
play soccer, 11p players play tennis, 8p
21. The price of a product manufactured at a players play golf and 2p players play
company KTM is given by the following none of the games. If p players play all of
formula: P = 6 – 0.03x, where P is the the games, how many players play only
price of a single product, and x is the one game?
number of products sold. What is the
maximum possible revenue for KTM? (1) The number of players who play
soccer and golf is half of the players who
A. 1000 play each of the rest two games
B. 600 (2) p = 3
C. 400
D. 300 A. Statement (1) ALONE is
E. 100 sufficient, but Statement (2)
ALONE is not sufficient
OVERLAPPING SETS B. Statement (2) ALONE is
sufficient, but Statement (1)
ALONE is not sufficient
22. 60% of the San Diego Zoo visitors are C. BOTH statements TOGETHER
single and all of the San Diego Zoo are sufficient, but NEITHER
family visitors have children. If 25% of statement ALONE is sufficient
families visiting the San Diego Zoo have D. EACH statement ALONE is
multiple children, what percentage of the sufficient
San Diego Zoo visitors have only one E. Statements (1) and (2)
child? TOGETHER are NOT sufficient
A. 15
B. 20 RATE
C. 30
D. 50 25. A bus from city M is traveling to city N
E. 75 at a constant speed while another bus is
making the same journey in the opposite
23. Out of 100 people surveyed, 60 were direction at the same constant speed.
women. If 10 were smoking women and They meet in point P after driving for 2
20 were smoking men, what percentage hours. The following day the buses do the
of men surveyed were non-smokers? return trip at the same constant speed.
One bus is delayed 24 minutes and the
A. 10 other leaves 36 minutes earlier. If they
B. 20 meet 24 miles from point P, what is the
C. 30 distance between the two cities?
D. 40
E. 50 A. 48
B. 72
C. 96
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D. 120 D. 4 hours
E. 192 E. 3 hours and 20 minutes
26. A train is traveling at a constant speed 29. Mac can finish a job in M days and Jack
and after making three one-hour stops can finish the same job in J days. After
reaches its destination. After waiting an working together for T days, Mac left and
hour it makes a return journey stopping a Jack alone worked to complete the
total of ten times, thirty minutes each but remaining work in R days. If Mac and
traveling at twice the speed. If both trips Jack completed an equal amount of work,
took the same amount of time, how many how many days would have it taken Jack
hours was the roundtrip? to complete the entire job working alone?
A. 14 (1) M = 20 days
B. 15 (2) R = 10 days
C. 16
D. 17 A. Statement (1) ALONE is
E. 18 sufficient, but Statement (2)
ALONE is not sufficient
27. A cook went to a market to buy some B. Statement (2) ALONE is
eggs and paid \$12. But since the eggs sufficient, but Statement (1)
were quite small, he talked the seller into ALONE is not sufficient
adding two more eggs, free of charge. As C. BOTH statements TOGETHER
the two eggs were added, the price per are sufficient, but NEITHER
dozen went down by a dollar. How many statement ALONE is sufficient
eggs did the cook bring home from the D. EACH statement ALONE is
market? sufficient
E. Statements (1) and (2)
A. 8 TOGETHER are NOT sufficient
B. 12
C. 15 30. Painters A and B can paint a house
D. 16 working alone in 20 and 30 days
E. 18 respectively. They started painting a
house together but then A left after a
WORK number of days but then rejoined B
before the job was completed. If B
28. It takes computer A 6 hours and 40 worked alone for 5 days and then A and
minutes to finish a job. If computer B can B together completed the work in 4 days,
process the same job in 10 hours, how after how many days of working
long will it take, for both computers together, did A leave B?
working together, to finish the job?
A. 4
A. 6 hours and 20 minutes B. 5
B. 5 hours and 10 minutes C. 6
C. 4 hours and 40 minutes D. 7
E. 8
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GEOMETRY
MIXTURE
34. A circle is inscribed in a half circle
31. How many liters of pure alcohol must be whose diameter is π. What is the ratio of
added to a 40-liter solution that is 10% the area of the half circle to the area not
alcohol by volume in order to double the covered by the inscribed circle?
alcohol proportion?
A. 1: 1
A. 4 B. 1: 2
B. 5 C. 1: 3
C. 10 D. 3: 4
D. 20 E. 2: 1
E. 40
35. If vertexes of a triangle are A (5, 0), B (x,
32. A Food and Drug lab has two new y) and C (25, 0), what is its area?
samples: a 240 gram cup of drip coffee,
which contains 124 mg of caffeine, and a (1) |x| = y = 10
60 gram cup of espresso, containing160 (2) x = |y| = 10
mg of caffeine. If a technician were to
create a new 120 gram cup sample that A. Statement (1) ALONE is
contained 50% coffee and 50% espresso, sufficient, but Statement (2)
how many mg of caffeine would the new ALONE is not sufficient
drink contain? B. Statement (2) ALONE is
sufficient, but Statement (1)
A. 111 ALONE is not sufficient
B. 121 C. BOTH statements TOGETHER
C. 144 are sufficient, but NEITHER
D. 191 statement ALONE is sufficient
E. 382 D. EACH statement ALONE is
sufficient
33. A kilogram of nut mixture contains X% E. Statements (1) and (2)
chestnuts and Y% walnuts and sells for TOGETHER are NOT sufficient
\$7.00/kg. If the ratio of chestnuts is
increased by 50% so that the new mixture
is sold for \$8.00/kg, what is the price of a 36. If an x meter long rope can enclose (or
kg of walnuts? cover) a maximum area of 50 square
meters, what is the approximate length of
A. \$1.00 the rope?
B. \$2.50
C. \$5.00 A. 4
D. \$7.50 B. 7
E. \$10.00 C. 10
D. 20
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## E. 25 probability that a contestant does not
taste all of the samples?
PROBABILITY
1
A.
37. A jar contains B blue balls, 6B + 10 12
yellow balls and 2B+5 green balls. What 5
B.
is the probability of taking out a blue or 14
green ball if there are no other balls? 4
C.
9
1 1
A. D.
5 2
1 2
B. E.
4 3
1
C.
3
1
D.
2
2
E.
3
## 38. A box has 6 red hats and 5 green hats.
What is the probability of drawing at
least one green hat in two consecutive
drawings if the hat is not replaced?
10
A.
11
8
B.
11
7
C.
12
5
D.
13
2
E.
7
## 39. At a blind taste competition a contestant
is offered 3 cups of each of the 3 samples
of tea in a random arrangement of 9
marked cups. If each contestant tastes 4
different cups of tea, what is the
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## C. BOTH statements TOGETHER
COMBINATIONS are sufficient, but NEITHER
statement ALONE is sufficient
40. Romi has a collection of 10 distinct D. EACH statement ALONE is
books (8 small and 2 large). In how many sufficient
ways can he select 5 books to take with E. Statements (1) and (2)
him on a trip if he has room for only 1 TOGETHER are NOT sufficient
large book?
A. 56
B. 126
C. 152
D. 196
E. 252
## 41. How many triangles and quadrilaterals
altogether can be formed using the
vertices of a 7-sided polygon?
A. 35
B. 40
C. 50
D. 65
E. 70
## 42. Set X has 5 integers: a, b, c, d, and e. If m
is the mean and D, where
( a −m ) 2 +(b − m ) 2 +( c − m ) 2 +( d −m ) 2 +( e− m ) 2
D= 5 , is the
standard deviation of the set X, is D>2 ?
## (1) a, b, c, d and e are different
integers
(2) m is an integer not equal to any
elements of the set X
## A. Statement (1) ALONE is
sufficient, but Statement (2)
ALONE is not sufficient
B. Statement (2) ALONE is
sufficient, but Statement (1)
ALONE is not sufficient
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ALGEBRA
A. |x| + |y| = 10
43. If 5x = y + 7, what is (x – y)? B. |x| > |y|
C. |x| > |y| +10
(1) xy = 6 D. |x| = |y|
(2) x and y are positive consecutive E. |x| - |y| = 5
integers
45. 5 boxes are placed in a stack by weight
A. Statement (1) ALONE is from lightest to heaviest. The heaviest
sufficient, but Statement (2) box weighs x kg and then next heaviest
ALONE is not sufficient weighs x% less than the heaviest box and
B. Statement (2) ALONE is the next heaviest box weighs x-x% less
sufficient, but Statement (1) than the next heaviest and so on. The
ALONE is not sufficient same ratio is maintained for all the boxes.
C. BOTH statements TOGETHER If the heaviest box weighs 10kg,
are sufficient, but NEITHER approximately what percent less weight
statement ALONE is sufficient is the lightest box than the heaviest one?
D. EACH statement ALONE is
sufficient A. 30
E. Statements (1) and (2) B. 40
TOGETHER are NOT sufficient C. 50
D. 60
44. x2 + y2 = 100. All of the following could E. 70
be true EXCEPT
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For explanations and detailed analysis, visit here:
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1 D 25 E
2 E 26 B
3 E 27 E
4 E 28 D
5 A 29 C
6 A 30 C
7 E 31 B
8 B 32 D
9 B 33 C
10 E 34 E
11 C 35 B
12 D 36 E
13 E 37 C
14 E 38 B
15 E 39 B
16 E 40 D
17 B 41 E
18 A 42 C
19 E 43 B
20 C 44 C
21 D 45 A
22 C
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GMAT Club Diagnostic Test | 5,572 | 17,302 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2019-30 | latest | en | 0.894291 |
http://www.netlib.org/lapack/lapack-3.1.1/manpages/man/manl/zgeevx.l | 1,555,924,050,000,000,000 | text/plain | crawl-data/CC-MAIN-2019-18/segments/1555578548241.22/warc/CC-MAIN-20190422075601-20190422101601-00237.warc.gz | 279,930,671 | 2,970 | .TH ZGEEVX 1 "November 2006" " LAPACK driver routine (version 3.1) " " LAPACK driver routine (version 3.1) " .SH NAME ZGEEVX - for an N-by-N complex nonsymmetric matrix A, the eigenvalues and, optionally, the left and/or right eigenvectors .SH SYNOPSIS .TP 19 SUBROUTINE ZGEEVX( BALANC, JOBVL, JOBVR, SENSE, N, A, LDA, W, VL, LDVL, VR, LDVR, ILO, IHI, SCALE, ABNRM, RCONDE, RCONDV, WORK, LWORK, RWORK, INFO ) .TP 19 .ti +4 CHARACTER BALANC, JOBVL, JOBVR, SENSE .TP 19 .ti +4 INTEGER IHI, ILO, INFO, LDA, LDVL, LDVR, LWORK, N .TP 19 .ti +4 DOUBLE PRECISION ABNRM .TP 19 .ti +4 DOUBLE PRECISION RCONDE( * ), RCONDV( * ), RWORK( * ), SCALE( * ) .TP 19 .ti +4 COMPLEX*16 A( LDA, * ), VL( LDVL, * ), VR( LDVR, * ), W( * ), WORK( * ) .SH PURPOSE ZGEEVX computes for an N-by-N complex nonsymmetric matrix A, the eigenvalues and, optionally, the left and/or right eigenvectors. Optionally also, it computes a balancing transformation to improve the conditioning of the eigenvalues and eigenvectors (ILO, IHI, SCALE, and ABNRM), reciprocal condition numbers for the eigenvalues (RCONDE), and reciprocal condition numbers for the right .br eigenvectors (RCONDV). .br The right eigenvector v(j) of A satisfies .br A * v(j) = lambda(j) * v(j) .br where lambda(j) is its eigenvalue. .br The left eigenvector u(j) of A satisfies .br u(j)**H * A = lambda(j) * u(j)**H .br where u(j)**H denotes the conjugate transpose of u(j). .br The computed eigenvectors are normalized to have Euclidean norm equal to 1 and largest component real. .br Balancing a matrix means permuting the rows and columns to make it more nearly upper triangular, and applying a diagonal similarity transformation D * A * D**(-1), where D is a diagonal matrix, to make its rows and columns closer in norm and the condition numbers of its eigenvalues and eigenvectors smaller. The computed reciprocal condition numbers correspond to the balanced matrix. Permuting rows and columns will not change the condition numbers (in exact arithmetic) but diagonal scaling will. For further explanation of balancing, see section 4.10.2 of the LAPACK Users\(aq Guide. .br .SH ARGUMENTS .TP 8 BALANC (input) CHARACTER*1 Indicates how the input matrix should be diagonally scaled and/or permuted to improve the conditioning of its eigenvalues. = \(aqN\(aq: Do not diagonally scale or permute; .br = \(aqP\(aq: Perform permutations to make the matrix more nearly upper triangular. Do not diagonally scale; = \(aqS\(aq: Diagonally scale the matrix, ie. replace A by D*A*D**(-1), where D is a diagonal matrix chosen to make the rows and columns of A more equal in norm. Do not permute; = \(aqB\(aq: Both diagonally scale and permute A. Computed reciprocal condition numbers will be for the matrix after balancing and/or permuting. Permuting does not change condition numbers (in exact arithmetic), but balancing does. .TP 8 JOBVL (input) CHARACTER*1 = \(aqN\(aq: left eigenvectors of A are not computed; .br = \(aqV\(aq: left eigenvectors of A are computed. If SENSE = \(aqE\(aq or \(aqB\(aq, JOBVL must = \(aqV\(aq. .TP 8 JOBVR (input) CHARACTER*1 = \(aqN\(aq: right eigenvectors of A are not computed; .br = \(aqV\(aq: right eigenvectors of A are computed. If SENSE = \(aqE\(aq or \(aqB\(aq, JOBVR must = \(aqV\(aq. .TP 8 SENSE (input) CHARACTER*1 Determines which reciprocal condition numbers are computed. = \(aqN\(aq: None are computed; .br = \(aqE\(aq: Computed for eigenvalues only; .br = \(aqV\(aq: Computed for right eigenvectors only; .br = \(aqB\(aq: Computed for eigenvalues and right eigenvectors. If SENSE = \(aqE\(aq or \(aqB\(aq, both left and right eigenvectors must also be computed (JOBVL = \(aqV\(aq and JOBVR = \(aqV\(aq). .TP 8 N (input) INTEGER The order of the matrix A. N >= 0. .TP 8 A (input/output) COMPLEX*16 array, dimension (LDA,N) On entry, the N-by-N matrix A. On exit, A has been overwritten. If JOBVL = \(aqV\(aq or JOBVR = \(aqV\(aq, A contains the Schur form of the balanced version of the matrix A. .TP 8 LDA (input) INTEGER The leading dimension of the array A. LDA >= max(1,N). .TP 8 W (output) COMPLEX*16 array, dimension (N) W contains the computed eigenvalues. .TP 8 VL (output) COMPLEX*16 array, dimension (LDVL,N) If JOBVL = \(aqV\(aq, the left eigenvectors u(j) are stored one after another in the columns of VL, in the same order as their eigenvalues. If JOBVL = \(aqN\(aq, VL is not referenced. u(j) = VL(:,j), the j-th column of VL. .TP 8 LDVL (input) INTEGER The leading dimension of the array VL. LDVL >= 1; if JOBVL = \(aqV\(aq, LDVL >= N. .TP 8 VR (output) COMPLEX*16 array, dimension (LDVR,N) If JOBVR = \(aqV\(aq, the right eigenvectors v(j) are stored one after another in the columns of VR, in the same order as their eigenvalues. If JOBVR = \(aqN\(aq, VR is not referenced. v(j) = VR(:,j), the j-th column of VR. .TP 8 LDVR (input) INTEGER The leading dimension of the array VR. LDVR >= 1; if JOBVR = \(aqV\(aq, LDVR >= N. .TP 8 ILO (output) INTEGER IHI (output) INTEGER ILO and IHI are integer values determined when A was balanced. The balanced A(i,j) = 0 if I > J and J = 1,...,ILO-1 or I = IHI+1,...,N. .TP 8 SCALE (output) DOUBLE PRECISION array, dimension (N) Details of the permutations and scaling factors applied when balancing A. If P(j) is the index of the row and column interchanged with row and column j, and D(j) is the scaling factor applied to row and column j, then SCALE(J) = P(J), for J = 1,...,ILO-1 = D(J), for J = ILO,...,IHI = P(J) for J = IHI+1,...,N. The order in which the interchanges are made is N to IHI+1, then 1 to ILO-1. .TP 8 ABNRM (output) DOUBLE PRECISION The one-norm of the balanced matrix (the maximum of the sum of absolute values of elements of any column). .TP 8 RCONDE (output) DOUBLE PRECISION array, dimension (N) RCONDE(j) is the reciprocal condition number of the j-th eigenvalue. .TP 8 RCONDV (output) DOUBLE PRECISION array, dimension (N) RCONDV(j) is the reciprocal condition number of the j-th right eigenvector. .TP 8 WORK (workspace/output) COMPLEX*16 array, dimension (MAX(1,LWORK)) On exit, if INFO = 0, WORK(1) returns the optimal LWORK. .TP 8 LWORK (input) INTEGER The dimension of the array WORK. If SENSE = \(aqN\(aq or \(aqE\(aq, LWORK >= max(1,2*N), and if SENSE = \(aqV\(aq or \(aqB\(aq, LWORK >= N*N+2*N. For good performance, LWORK must generally be larger. If LWORK = -1, then a workspace query is assumed; the routine only calculates the optimal size of the WORK array, returns this value as the first entry of the WORK array, and no error message related to LWORK is issued by XERBLA. .TP 8 RWORK (workspace) DOUBLE PRECISION array, dimension (2*N) .TP 8 INFO (output) INTEGER = 0: successful exit .br < 0: if INFO = -i, the i-th argument had an illegal value. .br > 0: if INFO = i, the QR algorithm failed to compute all the eigenvalues, and no eigenvectors or condition numbers have been computed; elements 1:ILO-1 and i+1:N of W contain eigenvalues which have converged. | 2,081 | 6,935 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2019-18 | latest | en | 0.635987 |
https://luschny.de/math/zeta/The-Bernoulli-Manifesto.html | 1,718,731,162,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861773.80/warc/CC-MAIN-20240618171806-20240618201806-00092.warc.gz | 331,183,620 | 26,319 | The Bernoulli numbers are the children of the zeta function ...
# The Bernoulli Manifesto
A survey on the occasion of the 300-th anniversary of the publication of
Jacob Bernoulli's Ars Conjectandi, 1713-2013.
Avant-propos: The starting point of this public letter was a discussion in the Newsgroup de.sci.mathematik. The central question was: Are the Bernoulli numbers misdefined? I presented the question in an open letter to Donald E. Knuth. Professor Knuth was kind enough to respond. Below is my response to Prof. Knuth. In order to understand this response the reading of the previous correspondence is recommended.
## The generating function approach
Dear Professor Knuth,
Thank you for taking the time to respond to my letter.
You seem to prefer the generating function $ze^{z}/(e^{z}-1)$ instead of $z/(e^{z}-1)$;
Yes and no. My primary concern is to find a substantive definition of the Bernoulli numbers. A definition that, if it is to be, even goes beyond the paradigmatic example given by Bernoulli himself and embeds naturally into the wider field of mathematics. Only then would I like to turn to the question of representations. And a generating function is a formal representation, that serves, first and foremost, simple computational manipulation of these numbers, but which should always remain subordinate to their content and never reach the rank of a substantive definition.
But to comply with the course of the discussion we will put the cart before the horse. What we can immediately agree upon is the generating function of the Bernoulli polynomials.
$$\frac{ze^{zx}}{e^{z}-1}=\sum \limits_{n=0}^{\infty}B_{n}(x)\frac{z^{n}}{n!} \ . \ \label{BP}$$
Taking this as the point of departure, only two things now come into consideration to be called Bernoulli numbers: the coefficients which result from inserting $x=0$ or $x=1$ into the generating function (\ref{BP}).
\begin{align} B^{(0)}(z) & = \frac{z}{e^{z}-1} \ = \, {\textstyle\sum\limits_{n=0}^{\infty}} B_{n}(0)\frac{z^{n}}{n!}\ \label{B0} \4pt] B^{(1)}(z) & = \frac{z}{1-e^{-z}} = % = \frac{ze^{z}}{e^{z}-1} {\textstyle\sum\limits_{n=0}^{\infty}} B_{n}(1)\frac{z^{n}}{n!}\ \label{B1} \end{align} So the question is which of the two sequences \begin{align} B_{n}(0) & = 1,-\frac{1}{2},\frac{1}{6},0,-\frac{1}{30},0,\frac{1}{42}, \ldots \\[4pt] B_{n}(1) & = 1,+\frac{1}{2},\frac{1}{6},0,-\frac{1}{30},0,\frac{1}{42}, \ldots \end{align} is more likely to represent the unity of mathematics (in the sense of B. Mazur's essay Bernoulli numbers and the unity of mathematics). Indeed, these two sequences differ only for n = 1 and also only because of the sign B_{1}(1)=-B_{1}(0). This is an unfortunate coincidence because it tends to belittle what is no bagatelle. To say that the sum of the coefficients of a polynomial has a certain value is a much more substantial statement than to give a name to its constant term. ### Bernoulli numbers and Stirling polynomials. … but I can think of many reasons to prefer the latter formula ... not least the connection with Stirling numbers in Concrete Math (7.52), although I realize that the former function does arise in my definition of Stirling polynomials in Concrete Math (6.50). Let us consider $$\left(B^{(1)}(z)\right)^{x} = x\sum_{n\geq0}\sigma_{n}(x)z^{n}\ .$$ You define the Stirling polynomials \sigma_{n} based on B^{(1)}(z), and this makes sense to me as thereby the relationship between Stirling polynomials and Bernoulli numbers can be easily expressed provided one sets \operatorname{B}_{n} = B_{n}(1). $$n!\sigma_{n}(1)=B_{n}(1) \quad(n\geq0) \ . \label{SPB}$$ With this choice \left(B^{(1)}(z)\right)^{m} for m=1 directly represents the relationship between the Stirling polynomials and the Bernoulli numbers. A simpler and more coherent relationship cannot be expected. Let us now consider the counterpart, the relation between \sigma_{n}(0) and B_{n}(0). It turns out a lot more problematic. In Concrete Math Exercise 6:18 we find the identity $$(x+1)\sigma_{n}(x+1)=(x-n)\sigma_{n}(x)+x\sigma_{n-1}(x)\ . \label{stirpoly}$$ Substituting x=0 we obtain \sigma_{n}(1) = -n \sigma_{n}(0). If we choose n = 0 then an arithmetical exception occurs because \sigma_{0}(x) = 1/x. Therefore \sigma_{0}(0) can not be B_{0}(0). If we choose n = 1 then unfortunately 1/2=-1/2 follows since \sigma_{1}(x)=1/2. Thus obviously the choice \operatorname{B}_{n} = B_{n}(0) has to be avoided here. In summary: If one chooses the definition \operatorname{B}_{n}=B_{n}(1), one can take instead of Concrete Math (7.44) \frac{z}{e^{z}-1} the expression \ B^{(1)}(z), Concrete Math (7.52) \frac{z}{1-e^{-z}} the expression \ B^{(1)}(z), Concrete Math (6.50) \frac{ze^{z}}{e^{z}-1} the expression \ B^{(1)}(z). This describes the connection between the Bernoulli numbers and the Stirling polynomials more clearly and uniformly. If we understand the search for the true Bernoulli numbers as a match, then identity (\ref{SPB}) scores \mathbf{1:0} for the choice \operatorname{B}_{n}=B_{n}(1). ### The Bernoulli function. Now let's look at the Bernoulli function \mathcal{B}(s). We write \tau = 2 \pi. $$\mathcal{B}(s)=-2\cos(s\tau/4)s!\,\zeta(s)/\tau^s \label{berfun}$$ The real Bernoulli function \mathcal{B}(s)=-s\zeta(1-s) … notice that there are many functions that interpolate the Bernoulli numbers at all positive integers; for example, multiply your Bernoulli function' by e^{2\pi is} or by \cos\pi s. Yes of course, there are also infinitely many functions which interpolate the numbers 1,1\cdot2,1\cdot2\cdot3,1\cdot2\cdot3\cdot4,…. Yet no one will come up with the idea to glue a cosine factor onto the \Gamma function found by Daniel Bernoulli and Leonhard Euler. And neither here. Of course it is true that, if we consider \mathcal{B}(s)\cos(\pi s), we could hit -1/2 at the point 1 by adding a few meaningless oscillations. But for heaven's sake, our definition is not chosen arbitrarily! The value of the definition lies in the connection with the Riemannian functional equation. In fact it is almost identical to its right hand side! $$\zeta(1-s) = 2\cos(s\tau/4)\,\Gamma(s)\,\zeta(s)/\tau^s \label{Z}$$ Using it we get the representation $$\mathcal{B}(s)=-s\zeta(1-s) \ . \label{BZ}$$ We introduce here basically a convenient notation which helps us to reap the fruits of the Riemann functional equation in the context of the Bernoulli numbers. The functional equation must stay intact of course. This approach can be continued to the representation of the Bernoulli polynomials by the Hurwitz zeta function. $$\zeta\left(-n,w\right) =-\frac{B_{n+1}(w)}{n+1}\qquad\left(n\geq0\right)$$ This is explained, for instance, in T. Apostol, Introduction to Analytic Number Theory, (th. 12.13 and 12.16). Setting w=1 we get $$B_{n}(1)=-n\zeta(1-n) \ . \label{bezeta}$$ Thus the Bernoulli function is exactly the confluent continuation of this identity. The cases n=0 and n=1 still require special consideration. In these cases \mathcal{B}(n)=\lim_{t\rightarrow n}\mathcal{B}(t) is to be understood. But both limits exist, \lim_{t\rightarrow0}\mathcal{B}(t)=1 and \lim_{t\rightarrow1}\mathcal{B}(t)=1/2. So once we have introduced the Bernoulli function \mathcal{B}(s) it makes sense to define \operatorname{B}_{n}=\mathcal{B}(n). And since $$\mathcal{B}(n)=B_{n}(1)$$ the score is \mathbf{2:0} for the choice \operatorname{B}_{n}=B_{n}(1). Only this choice correctly reflects the connection with the zeta function. However in contrast to the \zeta function the Bernoulli function does not have a pole at s=1. This fact makes it possible to determine the value of \operatorname {B}_{1} without having to rely on conventions. A more satisfying way to introduce the Bernoulli function directly without reference to the the \zeta function and its disruptive pole will be described below. It generalizes a representation of the Bernoulli numbers due to J. Worpitzky and was given by Helmut Hasse in 1930. ### The continuation of an almost miraculous formula. Consider the following beautiful formula $$\frac{\tau^{s}\mathcal{B}(s)}{s!}=-2\cos(s\tau/4)\,\zeta(s) \ \label{R}$$ ( \tau = 2 \pi), which is valid for all complex numbers if the function is continued in the points \left\{0,1\right\} by the respective limits. In particular we have $$\frac{\tau^{n}\operatorname{B}_{n}}{n!}=-2\cos(n\tau/4)\,\zeta(n)\quad (n \gt 1 \ \text{integer})\ .$$ Substituting in this formula \operatorname{B}_{1}=1/2 we obtain \pi=\pi, but substituting \operatorname{B}_{1}=-1/2 we obtain -\pi=\pi. Comparing (\ref{R}) with Concrete Math (6.89), a formula of Euler that you called an almost miraculous closed form for infinitely many infinite sums, $$\frac{\tau^{2n}B_{2n}(0)}{ \left( 2n \right) !} = -2(-1)^{n}\zeta(2n) \qquad(n \gt 0) \label{E}$$ one sees that (\ref{E}) is nothing other than the precursor of (\ref{R}), which in turn is nothing else than the Riemann functional equation (\ref{Z})! And (\ref{E}) is almost miraculous while (\ref{R}) might get a cosine factor attached? This is not possible. Setting \operatorname{B}_{1}=B_{1}(0) would compromise this highlight of mathematics and its history by implying -\pi=\pi. Match score \mathbf{3:0} for the choice \operatorname{B}_{n}=B_{n}(1) because it does not lead to such absurd consequences. An almost miraculous function, -\tau^{s}\zeta(1-s)/(s-1)! ### The functional equation of the Bernoulli function. Since 2 \cos(\pi s / 2) = i^s + (-i)^s and using the notation \tau = 2 \pi equation (\ref{R}) can also be written as $$\mathcal{B}(s) = - {s!} \ \zeta(s) \frac{i^s + (-i)^s }{\tau^s} . \label{bertau}$$ Furthermore, because of (\ref{BZ}) \mathcal{B}(1-s) = (s-1) \zeta(s) we obtain a self-referential representation of the \mathcal{B} function $$\mathcal{B}(s) = s! \, \frac{\mathcal{B}(1-s)}{1-s} \frac{i^s + (-i)^s }{\tau^s} .$$ The miraculous formula emerged as a functional equation of the Bernoulli function! This functional equation has also a symmetric variant like its \zeta-cousin: $$\left( \frac{s}{2} \right) ! \ \pi^{-s/2} \ \mathcal{B}(1-s) = \left( \frac{1-s}{2} \right)! \ \pi^{-(1-s)/2} \ \mathcal{B}(s) \label{symfe}$$ This means that the function on the left side of (\ref{symfe}) is unchanged by the substitution s \leftarrow 1-s . ### Some assorted facts about the Bernoulli function The Bernoulli function doesn't have any zero on the left of \Re(s) = \sigma = 0 (nor on the line \sigma = 0 — a fact which is indeed equivalent to the prime number theorem). The zeros split into those with \Im(\rho) = 0 and \Im(\rho) \neq 0. According to the Riemann Hypothesis the latter all lie on the line \sigma = 1/2. For those who try to find a counterexample to the Riemann hypotheses the Bernoulli function is a suitable tool because the Bernoulli function has the same zeros as the zeta function on this critical line. This follows from a fact discovered by Riemann: if \rho is a zero of the Riemann zeta function then so is 1−\rho. The zeta and the Bernoulli function on the critical line. The series expansion of \mathcal{B}(s) follows from the Laurent series expansion of the Riemann zeta function $$1 - \gamma s - \gamma_1 s^2 - \frac{\gamma_2 }{2}s^3 - \frac{\gamma_3 }{6} s^4 + O(s^5)$$ where \gamma is the Euler-Mascheroni constant and \gamma_n is the n-th Stieltjes constant. This fact can be used to define the Stieltjes constants as the coefficients of this expansion of the Bernoulli function. This definition is simpler than the definition usually given, namely as constants that occur in the Laurent series expansion of the Riemann zeta function. Thus we can write $$\operatorname{B}(s) \, = \, 1 - s \sum_{n \ge 0} \, \gamma_n \frac{s^n}{n!}.$$ For the controversial case s=1 we find \operatorname{B}_{1} =1 - 1/2 because it is well known that \sum_{n \ge 0} \gamma_n / n! = 1/2. This fact raises the score to \mathbf{4:0} for the choice \operatorname{B}_{n}=B_{n}(1). If we use the generalized Stieltjes constants \gamma_n(a) instead of the ordinary ones we arrive at the generalized Bernoulli function \operatorname{B}(s,a), which is a generalization analogous to the generalization of the Riemann Zeta function to the Hurwitz Zeta function. The Euler-MacLaurin summation can be used to prove the formula $$- \frac{\mathcal{B}'(0)}{\mathcal{B}(0)} = \gamma.$$ This means that - \gamma is the slope of the tangent line of the Bernoulli function at the origin — like the slope of the factorial function at this point! Julian Havil would be delighted. This makes one curious to gain more information about the function {- \mathcal{B}'(s)}/{\mathcal{B}(s)}, which can be written in terms of the zeta function as $$- \frac{\mathcal{B}'(s)}{\mathcal{B}(s)} = \frac{\zeta'(1-s)}{\zeta(1-s)} - \frac{1}{s}, \qquad (s \neq 0).$$ Lastly we mention that in the interval -2 \le s \le 1 $$\mathcal{B}(1-s) \approx \frac{1}{2} + \frac{5}{12}s + \frac{1}{12} s^2$$ is a fairly good first approximation as can be seen from the Euler-MacLaurin expansion of the \zeta function. The approximation is exact for s \in \{ -2, -1, 0, 1 \}. We will return to the Euler-MacLaurin expansion of the \zeta - function below. The function - \frac{\mathcal{B}'(s)}{\mathcal{B}(s)} hits Euler's \gamma at s=0. ### The Hadamard product of the Bernoulli function. The Bernoulli function can be written as a product \mathcal{B}(s) = T(s)R(s) where T(s) takes care of the zeros with \Im(\rho) = 0 and R(s) for those with \Im(\rho) \neq 0. (We will call T(s) the trivial part and R(s) the Riemann part of the Bernoulli function.) Let \gamma be Euler's constant and t = s/2-1/2 then T(s) is defined as $$T(s) = \sin(\pi t)\, \Gamma(t)\, e^{(2+\gamma)t } (2 \pi)^{-s}.$$ T(s) can be seen as a first approximation to the Bernoulli function which computes all values at odd integers exactly. The formula clearly displays the zeros at 3, 5, 7, ... (the vanishing of the Bernoulli numbers at these indices) due to the sine function term but also gives the value of \mathcal{B}(1) = 1/2 (understood as a limit). The second function, R(s), is quite a delicate function which Hadamard derived from the Weierstrass factorization theorem. It is a product over the zeros \rho of the Riemann zeta function with \Im(\rho) \neq 0. $$\mathcal{B}(s) = T(s) \prod_{\rho} \left(1-\frac{s}{\rho}\right) \exp \left( \frac{s}{\rho} \right) \qquad (s \gt 0)$$ The plot below shows the Bernoulli function, the trivial part of the Bernoulli function and an approximation using the Weierstrass factorization theorem where the product is taken over the first one hundred zeros of the zeta function. The Hadamard factorization of the Bernoulli function The fact that Weierstrass and Hadamard voted (implicitly) for \mathcal{B}(1) = 1/2 raises the match score to \mathbf{5:0}. Yet there is a second, simpler form of the factorization formula which highlights the value of the Bernoulli function at s=1 in a striking fashion. $$\mathcal{B}(s) = \frac{1}{2} \frac{\pi^{1/2-s/2}}{(1/2-s/2)!} \prod_{\rho} \left(1-\frac{s}{\rho}\right)$$ ### The alternating Bernoulli function. The alternating Riemann zeta function is defined as \zeta^{*}(s) = \zeta(s)(1- 2^{(1-s)}). The alternating Bernoulli function has the same definition as the Bernoulli function except that the zeta function is replaced by the alternating zeta function. $$\mathcal{B}^{*}(s) = -s \zeta^{*}(1-s). \label{betastardef}$$ For s=0 the right side of (\ref{betastardef}) is understood as a limit, \mathcal{B}^{*}(s) = \lim_{s \rightarrow 0} -s \zeta^{*}(1-s) = 0. In terms of the Bernoulli function the definition can be restated as $$\mathcal{B}^{*}(s) = \mathcal{B}(s) \left( 1-2^{s} \right). \label{betastar}$$ Using equation (\ref{bertau}) we can also express the alternating Bernoulli numbers in terms of the zeta function (using the notation \tau = 2 \pi) $$\mathcal{B}^{*}(s) = {s!}\ \zeta(s) \left(2^{s} -1 \right)\frac{ i^s + (-i)^s }{\tau^s} .$$ Introducing the function z_t(s) =(i/ ( t \pi))^s + (-i/(t \pi))^s we have the alternate form $$\frac{\mathcal{B}^{*}(s)}{s!} = \zeta(s) ( z_1(s)- z_2(s)).$$ ### The alternating Bernoulli numbers. The alternating Bernoulli numbers are the values of the alternating Bernoulli function at the nonnegative integers. $$\operatorname{B}^{*}_n = \mathcal{B}^{*}(n).$$ Like the Bernoulli numbers the alternating Bernoulli numbers are rational numbers. A distinctive feature characterizes them: reduced to lowest terms they all have the same denominator, which is 2. Thus it is convenient to introduce the integers \operatorname{G}_n = 2 \operatorname{B}^{*}_n . These numbers are known as the Genocchi numbers. $$\operatorname{G}_n \, = \, 0,\,−1,\,−1,\,0,\,1,\,0,\,−3,\,0,\,17,\,0,\,−155,\,0,\,2073, \ldots \quad (n \ge 0)$$ The Genocchi function 2 \mathcal{B}^{*}(x) = 2x\zeta(1-x)(2^x-1) It should be noted that our approach introducing the Genocchi numbers as \operatorname{G}_n = 2 \mathcal{B}^{*}(n) defines the numbers for all n \ge 0, unambiguously also in the case n=1. If we did not have the Bernoulli function as a fixed reference we would have to start another meaningless Glasperlenspiel (glass bead game) with generating functions, just running once again into similar problems as with the Bernoulli numbers. We would have to ask ourselves: is 2x/(1+e^x) or is -2x/(1+e^{-x}) the exponential generating function? Is \operatorname{G}_1 = 1 or \operatorname{G}_1 = -1? Match score \mathbf{6:0}. Herbert Wilf's famous dictum "A generating function is a clothesline on which we hang up a sequence of numbers for display" clearly implies that we first have to have the clothes which we wish to hang up — a simple fact often forgotten by users of generatingfunctionology. ### The Euler-MacLaurin expansion of the Riemann zeta function. The way the relationship between finite and infinite calculus is described in Concrete Math (2.6) also serves to illuminate the relationship between the zeta function and the Bernoulli numbers. So let \[ \zeta(s)=\lim_{N\rightarrow\infty}\sum_{n=1}^{N} \frac{1}{n^s}
If we expand analytically using the Euler-MacLaurin formula we find with $\operatorname{B}_{n}= B_{n}(1)$
$$\zeta(s)=\sum_{k=0}^{K}\frac{\operatorname{B}_{k}}{k!}s^{\overline{k-1}} + R(s,K) \ . \label{aztaknu}$$
$R(s,K)$ denotes a remainder term which is of no further interest here. The point is the closed form of the sum. Written explicitly:
\begin{align} \zeta(s) & = \frac{\operatorname{B}_{0}}{0!}s^{\overline{-1}}+\frac{\operatorname{B} _{1}}{1!}s^{\overline{0}}+\frac{\operatorname{B}_{2}}{2!}s^{\overline{1}} +\cdots+R(s,K)\nonumber\\ & = \frac{1}{s-1}+\frac{1}{2}+\frac{1}{12}s+\cdots+R(s,K) \nonumber \ . \end{align}
The Euler-MacLaurin formula, the calculation rules of rising powers, the Bernoulli numbers, everything blends in beautifully, as long as we select $\operatorname{B}_{1}=1/2$, and not $-1/2$. Match score $\mathbf{7:0}.$
And what about the practice? I pick out an example from a paper in the arXiv. It illustrates a situation which is typical for a whole class of Euler-MacLaurin expansions which by setting $\operatorname{B}_{1}=-1/2$ lose in formal content. The author gives the above expansion of the zeta function like this:
$$\zeta(s)=\frac{1}{s-1}+\frac{1}{2}+\sum_{k=1}^{K}\frac{\operatorname{B}_{2k}}{(2k)!} s(s+1)\cdots(s+2k-2)+R(s,K) \label{zetaexpl}$$
The clear and succinct formula (\ref{aztaknu}) has become a formulae mish-mash whose true structure still has to be reconstructed.
But the author (\ref{zetaexpl}) was virtually forced to choose the elaborated form, because only this allowed him to write $\operatorname{B}_{1}=1/2$ without coming into conflict with the convention $\operatorname{B}_{1}=-1/2$.
We see how the wrong choice of $\ \operatorname{B}_{1}$ can lead to a considerable loss of formal conciseness.
### The correct indexing of sums.
Dijkstra correctly said that most loops should run from 0 to n - 1, not from 1 to n. That's why the sum of $0^{m}+1^{m}+\cdots+(n-1)^{m}$ is cleaner (and more satisfying to a mature mathematician) than $1^{m}+\cdots+n^{m}$.
Agreed! But such a representation is with $\operatorname{B}_{1}=1/2$ just as possible as with $\operatorname{B}_{1}=-1/2$:
\begin{align*} {\textstyle\sum\limits_{k=0}^{n-1}}k^{m} & =\frac{1}{n+1}(B_{n+1}(m)-B_{n+1}(0))\ ,\\ {\textstyle\sum\limits_{k=0}^{n-1}} k^{m} & =\frac{1}{n+1}(B_{n+1}(m)-(-1)^{n+1}B_{n+1}(1)) \ . \end{align*}
As such, this point is not relevant to this discussion.
### Harmonic sum
But we will take up the advice. The harmonic sum is one of the recurring themes of Concrete Math. Let us consider formula (9.88), which is for $n\geq1$
$$\sum_{k=1}^{n}\frac{1}{k}=\ln(n)+\gamma+\frac{1}{2n}-\sum_{k=1}^{m} \frac{B_{2k}(0)}{2kn^{2k}}+R(2m,n)\ . \label{HMS1}$$
We counter with the formula $(n\geq1)$
$$\sum_{k=0}^{n-1}\frac{1}{k+1}=\ln(n+1)+\gamma-\sum_{k=1}^{m} \frac{B_{k}(1)}{k}(n+1)^{-k}+\tilde{R}(m,n)\;. \label{harm2}$$
First we note that both formulas provide the same: they approximate $H_{n}$, the $n$-th harmonic number, the second formula slightly better than the first. The right side of the second formula is shorter, the term $1/(2n)$ has disappeared, arguably it is more elegant and memorable.
What does this example teach us? It's worthwhile to follow the advice of Dijkstra and Knuth — but we always knew this — here about the correct way to index the sum on the left side of $(\ref{harm2}).$ But what happens if we were to write $B_{k}(0)$ on the right side of $(\ref{harm2})$? This neat formula would be broken!
### Bernoulli numbers and harmonic numbers.
It seems natural to mention next a representation of the Bernoulli numbers by the harmonic numbers.
$$\operatorname{B}_{n} = \sum_{k=0}^n \sum_{v=0}^k (-1)^v \binom{k}{v} H_{k+1}(v+2)^n \ . \label{harm}$$
I like this relationship well enough to award a point for it. This leads to the match score $\mathbf{8:0}$ as the right side of $(\ref{harm})$ reduces to $1/2$ in the case $n=1$.
This formula is the special case $x=1$ of the following representation of the Bernoulli polynomials:
$$B_{n}(x) = \sum_{k=0}^n \sum_{v=0}^k (-1)^v \binom{k}{v} H_{k+1}(x+v+1)^n$$
Further remarks and generalizations of these identities can be found on my blog Zeta Polynomials and Harmonic Numbers.
### Does the choice $B_1=1/2$ break widely-used formulas?
On the other hand the choice $B_{1}=1/2$ in our last example does not affect the first formula since the sum on the right side of $(\ref{HMS1})$ runs only over the even Bernoulli numbers. This observation is typical. One does not need to be afraid that classical formulas like the expansions of special functions, such as given for example in the Handbook of Mathematical Functions (HMF) or the Table of Integrals, Series and Products of I. S. Gradshteyn and I. M. Ryzhik would be affected by using $\operatorname{B}_{k} = B_{k}(1)$ and thus introduce confusion. This is not the case. A review of the relevant chapters of these two reference books shows that $\operatorname{B}_{1}$ is not used there but only the Bernoulli numbers at even indices.
Some function expansions in the typical cluttered $\operatorname{B}_{2k}$ style
(from H. Cohen, Number Theory)
In the first pages of Concrete Math you explain techniques that make summation user-friendly and comment that the omission of terms in a sum is not always a good idea:
But such temptations should be resisted; efficiency of computation is not the same as efficiency of understanding!
... and readability of formulas, one might add. Why can't the mathematicians trusted to remember the fact $\operatorname{B}_{k} = 0$ for odd $k\gt1$ when they really need it? Instead all this clutter!
Moreover the use of $\operatorname{B}_{k} = B_{k}(1)$ could have lead to simpler formulas also in the HMF, for example in (6.3.18) or (6.4.11). The classical Poincaré-type asymptotic expansion of the digamma function then could simply and concisely be written:
$$\psi(z) \sim \ln z - \sum_{k=1}^{\infty} \frac{\operatorname{B}_{k}}{k z^k} \label{digam}$$
This formula raises the match score to $\mathbf{9:0}$ and we note that $\operatorname{B}_{k} = B_{k}(1)$ gives us more freedom in the design of formulas.
### Non-natural ranges of validity.
The observation that some identities have a smaller and non-natural range of validity if one chooses $B^{(0)}(z)$ instead of $B^{(1)}(z)$ strikes me as a more serious problem. This is a sensitive seismograph indicating that some deeper tectonic layers are in conflict.
Let us denote the Euler numbers by $E_{n}$ and consider the following identity
$$\operatorname{B}_{n}=\genfrac{.}{.}{}{}{n}{4^{n}-2^{n}} \sum_{k=0}^{n-1} \binom{n-1}kE_{k} \ . \label{BE1}$$
The choice of $B^{(1)}(z)$ includes the case $n=1$ while the choice $B^{(0)}(z)$ precludes this case although no plausible reason suggests this.
For many authors, however, this is not enough. If the index $n=1$ is dropped, they exclude also all other odd $n$ from the sum. The result is the following disaster (dlmf.nist.gov/BP.4.15)
$$\operatorname{B}_{2n}=\genfrac{.}{.}{}{}{2n}{2^{2n}(2^{2n}-1)} \sum_{k=0}^{n-1}\binom{2n-1}{2k}E_{2k} \ . \label{BE2}$$
What was once a simple formula $(\ref{BE1})$ valid with $B^{(1)}(z)$ for all $n \gt 0$ now is a formula with a bloated notation $(\ref{BE2})$ and half as much content.
### Bernoulli numbers and Eulerian numbers.
Another non-natural limitation of the scope of validity can be observed when we look at the important connection between the Bernoulli numbers and the Eulerian numbers $\genfrac <> {0pt}{}{n}{k}$ (OEIS A173018, the Concrete Math version), which is given by the following two formulas:
$$\sum_{k=0}^n (-1)^k \genfrac <> {0pt}{}{n}{k} = \frac{4^{n+1}-2^{n+1}}{n+1} B_{n+1},$$ $$\sum_{k=0}^n (-1)^k \genfrac <> {0pt}{}{n}{k} {\binom{n}{k}}^{-1} = (n+1) B_n.$$
Both formulas are valid for $n \ge 0$ if we set $\operatorname{B}_{1} = 1/2$. Conversely, if we set $\operatorname{B}_{1} = -1/2$ then they are only valid for $n \ge 1$ and $n \ge 2$, respectively. Clearly the wrong definitions of the Bernoulli numbers limit our understanding of the situation and are a potential source of errors. This leads to the match score $\mathbf{10:0}$.
However, the most important relation between Bernoulli numbers and Eulerian numbers, assuming $\operatorname{B}_n = \operatorname{B}_n(1)$, is
$$\operatorname{B}_n = \sum_{k=0}^n\frac{(-1)^k}{k+1} \sum_{j=0}^k \genfrac <>{0pt}{}{n}{j} \binom{n-j}{n-k}.$$
The reason is the identity
$$(-1)^k \sum_{j=0}^k \genfrac <>{0pt}{}{n}{j} \binom{n-j}{n-k} = {\sum\limits_{j=0}^{k}}(-1)^{j}\binom{k}{j}\left(j+1\right)^{n} \ . \label{berulerian}$$
These are the Worpitzky numbers and we will meet them again under the name $\Delta^{k}\left({n}\right)$ in the analytic definition of the Bernoulli numbers below.
To round off these remarks: If we set $\operatorname{B}_n = \operatorname{B}_n(1)$ and let $A_n(x)$ denote the Eulerian polynomials, $E_n(x)$ the Euler polynomials and $\genfrac\{\}{0pt}{}{n}{k}$ the Stirling subset numbers, then the Bernoulli numbers also have their place in this remarkable chain of identities:
$A_{n}(-1)$ $= E_{n}(1) 2^n$ $= \zeta(-n)(2^{n+1}-4^{n+1})$ $= \operatorname{B}_{n+1} \frac{4^{n+1}-2^{n+1}}{n+1}$ $= \sum_{k=0}^n \genfrac\{\}{0pt}{}{n}{k} (-2)^{n-k} k!$ $= \sum_{k=0}^n \sum_{v=0}^k \binom{k}{v} (-1)^v 2^{n-k}(v+1)^n$
### Two identities.
Consider, for example, the fact that $$z/(e^{z}-1)-z/(e^{z}+1)=2z/(e^{2z}-1)\ , \label{identity1}$$ an identity with many more applications …
I'm not sure if I understand the meaning of this remark properly. If you feel that the existence of the identity $(\ref{identity1})$ is an argument in favor of the choice of the generating function $B^{(0)}(z)$, then I would point out that an identity of no less beauty exists which would then argue in favor of the choice of $B^{(1)}(z)$. Since
$$B^{(1)}(z)=ze^{z}/(e^{z}-1)=z/(1-e^{-z})$$
we are led by analogy to $(\ref{identity1})$ to the identity
$$z/(1-e^{-z})-z/(1+e^{-z})=2z/(e^{z}-e^{-z})\ . \label{identity2}$$
The identity $(\ref{identity2})$ does not change under the substitution $z \leftarrow -z$. Thus it can be written equivalently
$$z/(1+e^{z}) - z/(1-e^{z}) =2z/(e^{z}-e^{-z})\ .$$
### Topology and the Todd function.
Your last comment prompted me to reconsider the two rivaling generating functions $(\ref{B0})$ and $(\ref{B1})$ and to ask if there is something special which characterizes one of the two. And indeed there is:
Let $f(z)$ be a power series with constant term 1 such that the coefficient of $x^n$ in $(f(x))^{n+1}$ equals 1 for all n. (*)
As it has been observed by Friedrich Hirzebruch (Prospects in Mathematics, AM-70), $f(z) = z/(1-e^{-z})$ is the only power series satisfying (*). This function is called the Todd function (after John Arthur Todd).
Here we see the Bernoulli numbers show up in topology. Thus if we accept $B_1 = 1/2$ we can state: The Todd function is the exponential generating function of the Bernoulli numbers. And we can simply write
$\frac{z}{1-e^{-z}} = \sum_{k \ge 0} B_k \frac{x^k}{k!}.$
As we said in the introduction, before fixing the meaning of the Bernoulli numbers for the case $n=1$ it is a good idea to look not only at the paradigmatic case discussed by Bernoulli himself. Since the Todd class (defined by the Todd function) plays a fundamental role in the generalization of the theorem of Riemann-Roch for higher dimensions and in similar topological theorems we regard this as an strong argument in favor of $B_{1}=1/2$ and set the match score to $\mathbf{11:0}$.
### Von Staudt! Clausen! To the rescue!
B0 = 1 B1 = B2 = 1 − 1/2 − 1/3 B4 = 1 − 1/2 − 1/3 − 1/5 B6 = 1 − 1/2 − 1/3 − 1/7 B8 = 1 − 1/2 − 1/3 − 1/5 B10 = 1 − 1/2 − 1/3 − 1/11
Assume that the small table above is presented to you and you are asked to make a suggestion, spontaneous and intuitive, for the value of $\operatorname{B}_{1}.$ What would you say? $-1/2$? How would you justify that? Only $1-1/2$ comes to my mind. Admittedly this is not an argument.
However, two grand masters of the Bernoulli numbers, von Staudt and Clausen, have already answered the question back in 1840 when they independently found how the denominators of the $\operatorname{B}_{n}$ can be understood; their now famous theorem is only the phrasing of the above table in general terms.
I find this observation pretty enough to award a point for it. Match score $\mathbf{12:0}$ for von Staudt's and Clausen's voting for $\operatorname{B}_{n}=B_{n}(1).$
### The fabolous Fubini polynomials.
Let $n\brace k$ denote the Stirling set numbers. The Fubini polynomials are defined as $\operatorname{F}_n(x) = \sum_{k=0}^n (-1)^{n-k} k! \genfrac\{\}{0pt}{}{n}{k} x^k$ (see OEIS A278075).
Note that for all $n \ge 1 \$ $\operatorname{F}_n(0) = 0$ and $\operatorname{F}_n(1) = 1$. The Fubini polynomials have the remarkable property that $\int_{0}^{1} \operatorname{F}_n(x) \, dx = \operatorname{B_n}.$
$\int_{0}^{1}\ 1 \ dx = 1$ $\int_{0}^{1}\ x \ dx = \frac12$ $\int_{0}^{1}\ (2 x^2 - x) \ dx = \frac16$ $\int_{0}^{1}\ (6 x^3 - 6 x^2 + x) \ dx = 0$ $\int_{0}^{1}\ (24 x^4 - 36 x^3 + 14 x^2 - x) \ dx = -\frac{1}{30}$ $\int_{0}^{1}\ (120 x^5 - 240 x^4 + 150 x^3 - 30 x^2 + x) \ dx = 0$
There is hardly a more vivid representation of the Bernoulli numbers and we award a point for that. Match score $\mathbf{13:0}$.
### The Eulberian polynomials.
We admit it from the outset: the Eulberian polynomials do not exist in the literature and the name is fictitious. $$\operatorname{\beta}_{n}(x) = \frac{1}{n+1}\, \sum_{k=0}^n (-1)^k \frac{\genfrac <> {0pt}{}{n}{k}} {\binom{n}{k}} \, x^{n-k} \quad(n \ge 0).$$
Of course, $\operatorname{\beta}_{n}(1) = \operatorname{B}_{n}(1)$ as we have already seen above. What we want to emphasize here is: the Bernoulli numbers are not constants, they are the sum of n+1 terms. In other words, the coefficients of the polynomials form a regular triangle of rational numbers whose row sums are the Bernoulli numbers.
x0 x1 x2 x3 x4 x5
+ 1 · · · · · 1
0 + 1 / 2 · · · · 1 / 2
0 - 1/ 6 + 1 / 3 · · · 1 / 6
0 + 1 / 12 - 1 / 3 + 1 / 4 · · 0
0 - 1 / 20 + 11 / 30 - 11 / 20 + 1 / 5 · -1 / 30
0 + 1 / 30 - 13 / 30 + 11 / 10 - 13 / 15 + 1 / 6 0
Is it possible to find a (combinatorial) interpretation for these numbers? Perhaps building a bridge from the Bernoulli numbers to the Catalan numbers? It is worth a try.
For this purpose we introduce the following numbers generating a triangle that is the counterpart of Leibniz's harmonic triangle (which is generated when the falling factorial is used instead).
$$\operatorname{F}(n,k) \, = \, \frac{ (n+1)^{ \overline{n} } }{ k! \, (n-k)! }.$$
Then we can write
$$\frac{\operatorname{B}_n}{\operatorname{C}_n} \, = \, \sum_{k=0}^{n} (-1)^{k} \frac{\operatorname{E}(n, k)}{\operatorname{F}(n,k)},$$
where $\operatorname{C}_n$ denote the Catalan numbers and $\operatorname{E}(n,k)$ the Eulerian numbers in their CMath form A173018.
### The Akiyama-Tanigawa Algorithm.
For me, this is a proof without words. Match count $\mathbf{14:0}$.
### Bernoulli and Mona Lisa.
Todays long-standing mathematical conventions have many, many defects, and I can think of dozens of cases where changes would improve the current situation and make it easier on all future mathematicians.
But then we should do this! And we should do this refactoring, as it is called aptly by software engineers, with joy and without having a bad conscience.
But I would never think $\operatorname{B}_{1}$ to be anywhere near as important.
Well, it is difficult to respond to this argument. If Leonardo had decided to paint a big wart on the nose of Mona Lisa, it would also be not important. But it would hurt our aesthetic feelings. And so I perceive the Bernoulli bug: like an ugly wart on the Riemann functional equation of the zeta function.
We also have to contend with the broken windows syndrome: once shortcomings have crept in they will proliferate. Each bug confines our ability to understand the relationships. The non-natural limitation of the scope of validity of the connections between the Bernoulli numbers and other important numbers seen above is an example for this. It also means that a lot of work by mathematicians is invested to circumvent these shortcomings — effort and time that could be used for more creative thinking.
### The position of some experts
For this and many other reasons over nearly 40 years of working rather heavily with Bernoulli numbers, I have been convinced by the wisdom of the convention that has long been followed by the vast majority of the major writers on mathematics.
I admit that many writers in the 20th century followed the convention $\operatorname{B}_{1}=-1/2$. Among others two very influential books adopted this choice: Nörlund's Differenzenrechnung and the Handbook of Mathematical Functions. (Match score $\mathbf{14:2}$ ).
However, the so-called convention´ $\operatorname{B}_{1}=-1/2$ is by no means a uniformly accepted standard, and never was. For example in Whittaker and Watson's A Course of Modern Analysis, which was a standard reference and textbook in the first half of the 20-th century, the n-th Bernoulli number is $\mid B_{2n} \mid$.
Also Jean-Pierre Serre, the first Abel Prize winner, does not adhere to the HMF convention in his book A Course in Arithmetic, which stands on the recommendation lists of many universities. Serre sets, like Whittaker and Watson, for $n \geq 1 : \ \operatorname{B}_{n}=|B_{2n}(0)| = |B_{2n}(1)|.$ So do you think Serre confuses the current generation of mathematicians? I do not think so since even a simple student of mathematics like me was able to handle his definition.
My impression is that the number of modern authors who use $\operatorname{B}_{1}=1/2$ is steadily increasing. In publications which range from the popular book by Conway and Guy, The book of numbers to the much used and valued textbook by Jürgen Neukirch, Algebraic Number Theory. Neukirch defines the Bernoulli numbers via the series expansion of $\ F(t)\, =\, B^{(1)}(t) \$ $(\ref{B1})$ and writes (p. 427f):
The relation of the Bernoulli numbers to the zeta function gives them a special arithmetic significance. The first Bernoulli numbers are $B_{0}=1,\,B_{1}=\frac{1}{2},\,B_{2}=\frac{1}{6},\,B_{3}=0,\, B_{4} =-\frac{1}{30},\,B_{5}=0,\,B_{6}=\frac{1}{42}.$ In the classical literature, it is usually the function $t/(e^t-1)$, but the above definition is more natural and better suited for the further development of the theory.
In the meantime, the monograph of T. Arakawa, T. Ibukiyama, and M. Kaneko, Bernoulli numbers and zeta functions, Springer 2014, has been published, which is based on the definition $B_{1} = \frac{1}{2}$.
### Generalized Bernoulli numbers.
To explain the last remark of Neukirch a little bit we look at the generalized Bernoulli numbers $B_{n,\chi}$ as defined in analytic number theory. Let $\chi$ be a Dirichlet character and $f_{\chi}$ the conductor of $\chi$. Then
$$\sum_{n\ge 0} B_{n,\chi} \frac{t^n}{n!} := \sum_{j=1}^f \chi(j) \frac {t e^{jt}}{e^{f t}-1} .$$ When $\chi = 1$, we have $$\sum_{n\ge 0} B_{n,1} \frac{t^n}{n!} = \frac{t e^t}{e^t - 1} = \frac{t}{1-e^{-t}}$$ Thus $B_{n,1} = B_n$ (which is a natural request) if and only if $B_1$ is set to $1/2$. (Match score $\mathbf{15:2}).$
### How Terence Tao introduces the Bernoulli numbers.
The way Terence Tao introduces the Bernoulli numbers in his blog The Euler-Maclaurin formula, Bernoulli numbers, the zeta function, and real-variable analytic continuation highlights how naturally the right choice of the recurrence leads to the generating function of the Bernoulli numbers. Therefore we cite him at length:
"We define the Bernoulli numbers $B_0, B_1, \ldots$ recursively by the formula
$$\sum_{k=0}^{s-1} \binom{s}{k} B_k = s \label{taodef}$$
for all $s = 1,2,\ldots$, or equivalently
$$B_{s-1} := 1 - \frac{s-1}{2} B_{s-2} - \frac{(s-1)(s-2)}{3!} B_{s-3} - \ldots - \frac{1}{s} B_0.$$
The first few values of $B_s$ can then be computed:
$B_0=1;\ B_1=1/2;\ B_2=1/6;\ B_3=0;\ B_4=-1/30;\ \ldots.$
From $(\ref{taodef})$ we see that
$$\sum_{k=0}^\infty \frac{B_k}{k!} [ P^{(k)}(1) - P^{(k)}(0) ] = P'(1) \label{taopoly}$$
for any polynomial $P$ (with ${P^{(k)}}$ being the ${k}$-fold derivative of $P$); indeed, $(\ref{taodef})$ is precisely this identity with $P(x) := x^s$, and the general case then follows by linearity. "
Match score $\mathbf{16:2}$ since $(\ref{taopoly})$ is the nucleus of the Euler-Maclaurin formula and does not hold if $B_1=-1/2$. Tao then continues:
"As $(\ref{taopoly})$ holds for all polynomials, it also holds for all formal power series (if we ignore convergence issues). If we then replace $P$ by the formal power series
$$P(x) = e^{tx} = \sum_{k=0}^\infty t^k \frac{x^k}{k!}$$
we conclude the formal power series (in $t$) identity
$$\sum_{k=0}^\infty \frac{B_k}{k!} t^k (e^t-1) = t e^t$$
leading to the familiar generating function
$$\sum_{k=0}^\infty \frac{B_k}{k!} t^k = \frac{t e^t}{e^t-1}$$
for the Bernoulli numbers."
Tao's 'familiar generating function' is of course the same as our $(\ref{B1}) \ \frac{t}{{1-e^{-t}}}$.
After seeing a 21st century mathematician at work we now go back three centuries to interview two other important mathematicians: Sansei Takekazu-Kowa Seki and Jacob Bernoulli. It would be unfair not to seek their opinion. After all they invented the 'Bernoulli numbers' as they were christened by Leonhard Euler.
### Sansei Takekazu-Kowa Seki's sums of equal powers.
In 1712, one year before the appearance of Bernoulli's posthumously published Ars Conjectandi, the Japanese mathematician Seki introduced the Bernoulli numbers in his work Katsuyo Sampo while developing a method to calculate sums of powers.
He called his method Ruisai Shosa-ho and explained it with reference to the following table (reproduced here shortened).
1 2 3 4 5 6 7 8 9 x1 1 (1) x2 1 2 (1) x3 1 3 3 (1) x4 1 4 6 4 (1) x5 1 5 10 10 5 (1) x6 1 6 15 20 15 6 (1) x7 1 7 21 35 35 21 7 (1) x8 1 8 28 56 70 56 28 8 (1) σn-1 1 1/2 1/6 0 −1/30 0 1/42 0 1/30
This is obviously the binomial triangle where the bottom bar shows the Seki numbers $\sigma_n,$ as the Bernoulli numbers could also be called for good historic reason. We learn from this computational scheme that the Seki numbers start $1,1/2, \ldots$. (Match score $\mathbf{17:2}.$ ) Seki used this table to calculate the sum of powers according to the formula:
$$\sum_{k=1}^nk^{m-1} = \frac 1m\sum_{k=0}^{m-1}\sigma _k\binom mkn^{m-k}\qquad (n \gt 0, m \gt 0).$$
### And Jacob, what did you say?
Summae Potestatum $\smallint n = \frac12 nn + \frac12 n \\ \smallint nn \ =\ \frac13 n^3 + \frac12 nn + \frac16 n \\ \smallint n^3 = \frac14 n^4 + \frac12 n^3 + \frac14 nn \\ \smallint n^4 = \frac15 n^5 + \frac12 n^4 + \frac13 n^3 - \frac{1}{30}n \\ \smallint n^5 = \frac16 n^6 + \frac12 n^5 + \frac{5}{12} n^4 - \frac{1}{12} nn \\ \smallint n^6 = \frac17 n^7 + \frac12 n^6 + \frac12 n^5 - \frac16 n^3 + \frac{1}{42} n \\ \smallint n^7 = \frac18 n^8 + \frac12 n^7 + \frac{7}{12} n^6 - \frac{7}{24}n^4 + \frac{1}{12} nn \\ \smallint n^8 = \frac19 n^9 + \frac12 n^8 + \frac23 n^7 - \frac{7}{15}n^5 + \frac29n^3 - \frac{1}{30} n \\ \smallint n^9 = \frac{1}{10} n^{10} + \frac{1}{2} n^9 + \frac{3}{4} n^8 - \frac{7}{10} n^6 + \frac{1}{2} n^4 - \frac{1}{12} nn \\ \smallint n^{10} = \frac{1}{11} n^{11} + \frac12 n^{10} + \frac56 n^9 - 1 n^7 + 1 n^5 - \frac12 n^3 + \frac{5}{66} n \\$
Quin imó qui legem progressionis inibi attentuis ensperexit, eundem etiam continuare poterit absque his ratiociniorum ambabimus : Sumtâ enim $c$ pro potestatis cujuslibet exponente, fit summa omnium $n^c$ seu
$\int n^c = \frac{1}{c + 1}n^{c+1} + \frac12 n^c + \frac{c}{2} An^{c-1} + \frac{c \cdot c - 1 \cdot c - 2}{2 \cdot 3 \cdot 4} Bn^{c-3}$$+\frac{c \cdot c - 1 \cdot c - 2 \cdot c - 3 \cdot c - 4}{ 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6} C n^{c-5}$$+ \frac{c \cdot c - 1 \cdot c - 2 \cdot c - 3 \cdot c - 4 \cdot c - 5 \cdot c - 6 }{ 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 } Dn^{c-7} \cdots \text{& ita deinceps,}$
exponentem potestatis ipsius n continué minuendo binario, quosque perveniatur ad $n$ vel $nn$. Literae capitales A, B, C, D & c. ordine denotant coëfficientes ultimorum terminorum pro $\smallint nn,\ \smallint n^4,\ \smallint n^6,\ \smallint n^8$ & c. nempe
$A = \frac16 , B = -\frac{1}{30}, C =\frac{1}{42}, D = -\frac{1}{30} \ .$
### Summae Potestatum.
The last section above reproduces Jacob's exposition on computing sums of equal powers, as they were published in Basel in 1713, eight years after Bernoulli's death, in his book Ars Conjectandi (in the second part, on pages 97-98). We reconstruct the formula in the notation of Concrete Math, with the factorial and the falling factorial powers, as they are defined there on page 47 and 52.
\begin{align*} x^{\underline{m}} & = {\displaystyle\prod_{0\leq k\leq m-1}} \left( x-k\right) \, \quad \left( m\geq0\right) \ ; \\ x^{\underline{m}} & ={\displaystyle\prod_{m\leq k\leq-1}} \left( x-k\right)^{-1} \ \ \left( m \lt 0\right) \ . \end{align*}
Bernoulli does not use the notation $\operatorname{B}_{k},$ but writes the constants either explicitly or uses the symbols A, B, C, D, …. These constants are in a one-to-one relationship with the constants $\operatorname{B}_{k}$, as can be seen from Bernoulli's formula:
$\operatorname{B}_{0}=1,\,\operatorname{B}_{1}=\frac{1}{2},\,\operatorname{B}_{2}=A=\frac{1}{6},\,\operatorname{B}_{3}=0,\,\operatorname{B}_{4}=B=-\frac{1}{30},$$\operatorname{B}_{5}=0,\,\operatorname{B}_{6}=C=\frac{1}{42},\,\operatorname{B}_{7}=0,\,\operatorname{B}_{8}=D=-\frac{1}{30}, \ldots$
This yields Bernoulli's Summae Potestatum in the following form:
\begin{align} \int n^{c} & = \frac{1}{c+1}\operatorname{B}_{0}n^{c+1}+\operatorname{B}_{1}n^{c}+\frac{c}{2}\operatorname{B}_{2}n^{c-1} +\frac{c(c-1)(c-2)}{24}B_{4}n^{c-3}+ \ldots \\ & = \sum_{k\geq0}\frac{\operatorname{B}_{k}}{k!}c^{\underline{k-1}}n^{c-k+1} \end{align}
If there is anything that deserves the name of Bernoulli numbers, these numbers are it. And although Bernoulli did not use a special notation for $\operatorname{B}_{0}$ and $\operatorname{B}_{1}$ he did use the value $\operatorname{B}_{1}=\frac{1}{2}$. (Match score $\mathbf{18:2}.$ )
And every mathematician, be it a major or a minor one who wants to change this, please stand up and explain it to the mathematical community. However, those who have changed it without giving an explanation have been guilty of counterfeiting in a particularly severe case.
### Who is to blame for the confusion?
... the amount of confusion that a change would produce is vast. I find your comment that "the confusion' would be small" the understatement of the century.
It's still a bit early in this century to decide how big this compliment really is ;-).
A look at the history of the Bernoulli numbers does not let me share this view. Earlier mathematicians such as Francois de Viete, Abraham de Moivre or Leonhard Euler referred in their papers to the Bernoulli numbers as defined by Bernoulli.
Euler's summation formula, as written by Euler.
(from D. J. Pengelley, Dances between continous and discrete.)
Thus a change was made sometimes after Bernoulli. I do not know who was the first to change it and when it happened. Nor have I heard anyone explain the reason for this change. On the other hand, the choice of $B_n = B_n(1)$ is much more common than it seems that you assume as some outstanding mathematicians today use it in their textbooks and in their lectures, as we saw above.
So if there is a confusion, then it is already here. We have seen the example of Wikipedia's wrong definition of the Todd function and we have seen the dangers hidden in the non-natural ranges of validity of identities for Euler and Eulerian numbers if the wrong definition is chosen. So the question is only: How do we deal with the confusion?
And no, it is not too late´ to change. Mathematicians are well prepared to handle it. To learn how to live with the risks of conventions and notations belongs to the professional training of each mathematician. This may reassure us.
### What can we do?
... and what should we not do?
For instance we could introduce new names to disambiguate the two possible choices of $B_n$. In fact this has been done on the English Wikipedia and the nomenclature 'first Bernoulli numbers' and 'second Bernoulli numbers' was introduced. This disservice reflects the maximum credible misunderstanding. There is of course only one kind of Bernoulli numbers $B_{n}$.
This scholasticistic nonsense culminates on the German Wikipedia. Not only the notation $B_n^{*}$ is introduced for what is now called "Bernoulli-Zahlen zweiter Art" but the two 'kinds' are also mixed in identities like
$$\sum_{i = m}^{n} f(i) = \sum_{j = 0}^{\infty} \frac{1}{j!} \left(B_{j}^{\ast} f^{(j-1)}(n) - B_{j} f^{(j-1)}(m)\right) .$$
We can only hope that some insightful editors soon put an end to this moonshine.
### Is there a canonical representation of the Bernoulli numbers?
Why is there still no canonical representation of the Bernoulli numbers? even though they are so central as Mazur writes.
… Bernoulli numbers sit in the center of a number of mathematical fields, and whenever, for a given index $k$ the Bernoulli number $\operatorname{B}_{k}$ exhibits some particular behavior, these different mathematical fields seem to feel the consequences, each in their own way. (B. Mazur, Bernoulli numbers and the unity of mathematics.)
It is this question which should really worry us because it points to a theoretical deficit, a lack of understanding.
It is my belief that this can be overcome. Consider how Euler proceeded in the similar case of the factorial numbers: He has introduced a definition, that of the gamma function, which to this day is the canonical way to interpolate the factorial numbers because it has more advantages than possible alternative definitions.
This approach characterizes progress at the highest mathematical level, which is the level of definitions. If a definition is successful it makes conventions superfluous. This objective can be achieved here as well — by introducing the Bernoulli function. Below we will give an analytic definition independent from the zeta function. This will be our substantive definition, as we called it above, in contrast to the controversial formal definitions via generating functions.
## The analytical definition of the Bernoulli numbers.
Definition. The Bernoulli numbers are the values of the function $\mathcal{B}(s)$ on the non-negative integers. Here
$$\mathcal{B}(s) = \sum\limits_{k=0}^{\infty}\frac{\Delta^{k}\left({s}\right)}{k+1} \quad \text{ where} \quad \Delta^{k}\left({s}\right) = {\sum\limits_{v=0}^{k}}(-1)^{v}\binom{k}{v}\left(v+1\right)^{s} \ . \quad \label{berndef}$$
In the definition ($\ref{berndef}$) we use the notation $\mathcal{B}(s)$ like we used it in the definition of the Bernoulli function $\mathcal{B}(s)=-s\zeta(1-s).$ How does that fit together?
Well, a theorem of Helmut Hasse states that this is one and the same! Hasse gives an elementary, purely real-analytic proof of this identity.
By interpreting the variable $s$ over the natural numbers, $n\leftarrow s$, and adapting the summation limit $n\leftarrow\infty$, we obtain a classic representation of the Bernoulli numbers, the Worpitzky representation (J. Worpitzky, Studien über die Bernoullischen und Eulerschen Zahlen, Crelle 94 (1883), formula (36)).
$$\operatorname{B}_{n}=\sum_{k=0}^{n}\frac{\Delta^{k}(n)}{k+1} \ .$$
Can an embedding of discrete identities into continuous mathematics be more natural? Should we not maintain such a relationship, build on it and remove conventional obstacles from the path?
$B_{0}$ $1$ $B_{1}$ $1/1 -1/2$ $B_{2}$ $1/1 -3/2 +2/3$ $B_{3}$ $1/1 -7/2 +12/3 -6/4$ $B_{4}$ $1/1 -15/2 +50/3 -60/4 +24/5$ $B_{5}$ $1/1 -31/2 +180/3 -390/4 +360/5 -120/6$ $B_{6}$ $1/1 -63/2 +602/3 -2100/4 +3360/5 -2520/6 +720/7$
Definition. Bernoulli polynomials are the values of the function $\mathcal{B}(s,x)$ at the non-negative integers $s$. Let $x$ be a real number $\gt -1$ and
$$\mathcal{B}(s,x) = \sum\limits_{k=0}^{\infty} \frac{\Delta^{k}\left(s,x \right)}{k+1} \quad \text{where } \quad \Delta^{k}\left(s,x\right) = \sum\limits_{v=0}^{k}(-1)^{v}\binom{k}{v}\left(x+v+1\right)^{s} \ . \label{bernpolydef}$$
Hasse's theorem in its general form states
$$\mathcal{B}(s,x) = -s \zeta(1-s,x) \ .$$
$\zeta(s,x)$ is the Hurwitz zeta function. Thus we obtain for $n \gt 0$ and real $x$
$$\operatorname{B}_{n}(x) = -n \zeta(1-n,x) \ .$$
### The formal proof
The infinite series $\sum_{v=0}^{\infty}\frac{1}{n+1} \Delta_n ( {v^s} )$ converges for all complex $s$ and represents the entire function $- s\zeta(1-s)$, the Bernoulli function.
We quote Hasse's proof; the translation is ours.
(I) The series $\sum_{v=0}^{\infty}\frac{1}{n+1} \Delta_n \left( \frac{1}{v^s} \right)$ converges for $\Re(s) \gt 0$ and equals there $s\zeta(s+1)$. This results from the following formal computation:
\begin{align*} \frac{1}{v^s} &= \frac{1}{\Gamma(s)}\int_0^{\infty} e^{-vt}t^{s-1}dt, \6pt] \Delta_n \left(\frac{1}{v^s} \right) &= \sum_{v=0}^{n} (-1)^v \binom{n}{v}\frac{1}{(v+1)^s} \\[4pt] &= \frac{1}{\Gamma(s)}\int_0^{\infty} \sum_{v=0}^{\infty} (-1)^v \binom{n}{v} e^{-(v+1)t} t^{s-1} dt \\[4pt] &= \frac{1}{\Gamma(s)}\int_0^{\infty} \sum_{v=0}^{\infty} (1-e^{-t})^n e^{-t} t^{s-1} dt, \\[6pt] \sum_{v=0}^{\infty}\frac{1}{n+1} \Delta_n \left( \frac{1}{v^s} \right) &= \frac{1}{\Gamma(s)} \sum_{v=0}^{\infty} \int_0^{\infty} \frac{(1-e^{-t})^n}{n+1} e^{-t} t^{s-1} dt \\[4pt] &= \frac{1}{\Gamma(s)} \int_0^{\infty} \sum_{v=0}^{\infty} \frac{(1-e^{-t})^n}{n+1} e^{-t} t^{s-1} dt \qquad (*) \\[4pt] &= \frac{1}{\Gamma(s)} \int_0^{\infty} \frac{t}{1-e^{-t}}e^{-t}t^{s-1} dt \\[4pt] &= \frac{1}{\Gamma(s)} \int_0^{\infty} \sum_{n=1}^{\infty} e^{-nt} t^{s} dt \\[4pt] &= \frac{1}{\Gamma(s)} \sum_{n=1}^{\infty} \int_0^{\infty} e^{-nt} t^{s} dt \qquad \qquad \qquad \quad (**) \\[4pt] &= s \sum_{n=1}^{\infty} \frac{1}{\Gamma(s+1)} \int_0^{\infty}e^{-nt} t^{s} dt \\[4pt] &= s \sum_{n=1}^{\infty} \frac{1}{n^{s+1}} \\[4pt] &= s \zeta(s+1), \end{align*} To justify this only the commutativity of summation and integration of (*) and (**) is to be shown. But this follows easily from the uniform convergence of the two sums in question \[ \sum_{v=0}^{\infty} \frac{(1-e^{-t})^n}{n+1} e^{-t} t^{\sigma-1} dt \quad \text{and} \quad \sum_{n=1}^{\infty} e^{-nt} t^{\sigma} \qquad (\sigma=\Re(s) \gt 0)
which holds throughout the integration interval $0 \le t \le \infty$.
We use the notation $\vec{x}_n= x_1+\ldots+x_n$ and $s^{\overline{n}} = s(s+1) \ldots (s+n-1)$.
(II) The series $\sum_{v=0}^{\infty}\frac{1}{n+1} \Delta_n \left( \frac{1}{v^s} \right)$ also converges for $\Re(s) \le 0$, uniformly in any bounded region of this half-plane. It represents the analytic continuation of $s\zeta(1+s)$ in the left half-plane.
As the starting point for the proof I choose the integral representation of the initial terms of the difference series, which is easly shown: \begin{align*} \Delta_n \left( \frac{1}{v^s} \right) & = s^{\overline{n}}\int_0^1 \ldots \int_0^1 \frac{1}{(1+\vec{x}_n)^{n+s}} dx_1 \ldots dx_n \4pt] & = s^{\overline{n}}\int_0^1 \ldots \int_0^1 \frac{(1+\vec{x}_n)^{-s+p}}{(1+\vec{x}_n)^{n+p}} dx_1 \ldots dx_n \end{align*} Here p is an arbitrary fixed real positive number which is not an integer. For \sigma = \Re(s) \le0 we get the estimate \begin{align*} \left| \Delta_n \left( \frac{1}{v^s} \right)\right| & \leq \left| s^{\overline{n}} \right|(n+1)^{-\sigma+p} \int_0^1 \ldots \int_0^1 \frac{1}{(1+\vec{x}_n)^{n+p}} dx_1 \ldots dx_n \\ & = \left| \frac{s^{\overline{n}}}{p^{\overline{n}}} \right| (n+1)^{-\sigma+p} \Delta_n \left( \frac{1}{v^p} \right). \end{align*} \begin{equation*} \text{It is well known that } \frac{1}{\Gamma(s)} \sim \frac{s^{\overline{n}}}{n! \, n^{s-1}}. \text{ Thus } s^{\overline{n}} \sim n! \, n^{s-1} \frac{1}{\Gamma(s)} \end{equation*} uniformly in any bounded region of the s-plane. Therefore \begin{align*} \left| \Delta_n \left( \frac{1}{v^s} \right) \right| \lesssim \frac{n! \, n^{\sigma-1} \left| \frac{1}{\Gamma(s)} \right| }{ n! \, n^{p-1} \ \frac{1}{\Gamma(p)} } n^{-\sigma+p} \, \Delta_n \left( \frac{1}{v^p} \right) = \frac{\Gamma(p)}{|\Gamma(s)|} \Delta_n \left( \frac{1}{v^p} \right) \end{align*} uniformly in any bounded region of the s-plane. (II) now follows from (I). Helmut Hasse, Ein Summierungsverfahren für die Riemannsche \zeta-Reihe, Math. Z. 32 (1930). The paper can be freely obtained from the Center for Retrospective Digitization in Göttingen. ### But no, no, ten thousand times no! This was your response in Two Notes on Notation on the view that one should leave 0^{0} undefined. And it is my response to your view, that we should adopt the convention \operatorname{B}_{1} = -1/2. I believe in the unity of mathematics as described by Barry Mazur and take Leonhard Euler as my advisor how to achieve such a unity: by regarding the problem as an interpolation problem which generalizes a certain sequence of rational numbers to an entire function in the complex plane. In any case mathematics is not a bunch of disparate conventions which must be maintained in its present form just to protect some mathematicians from a possible confusion. And I would never dare to stick a number on the name of Jacob Bernoulli knowing that he himself never used it in this form. And is there a need for a revision of Bernoulli's insights on this topic? Not at all. The embedding of the Bernoulli numbers via the Worpitzky representation into the reflected zeta function, as has been shown by Helmut Hasse, is the most convincing argument. Let us use the function \[ \mathcal{B}(s) = \displaystyle\sum\limits_{k=0}^{\infty}\frac{\Delta^{k}\left({s}\right)}{k+1}
to the same beneficial effect for the Bernoulli numbers as the Euler gamma function showed for the factorial numbers. By this the Bernoulli numbers will find their natural place in mathematics, free from all conventions.
Dear Professor Knuth, thank you very much for your reply. I would like to inform you that I have posted your answer, as announced, in the newsgroup de.sci.mathematik. Since I was accused to waste your precious time with such a trivial question, I do not send you this answer directly, but leave it as my Bernoulli Manifesto on the Internet.
Sincerely,
Peter Luschny
The Bernoulli Manifesto. Copyright © 2013 by Peter Luschny. This manifesto was written on the occasion of the 300-th anniversary of the publication of Jacob Bernoulli's 'Ars Conjectandi'. Please feel free to share The Bernoulli Manifesto, which is available under the Creative Commons Attribution 3.0 Unported License. Help Jacob Bernoulli to regain the power over his numbers in the public discourse.
### Update 2020
An alternative systematic approach is described in my paper An Introduction to the Bernoulli function, arXiv:2009.06743. A HTML-version of this paper is here.
### Update 2021
Donald Knuth: Concrete Mathematics and Bernoulli
(from Don Knuth's homepage.)
After the many years I have stood up for this view, I was very touched by this news. But I was even more moved by the fact that and how Knuth reacted.
After Russ Cox showed the above screenshot on Twitter, several Twitter-typical comments came in, which I would like to put at the end. Because in part they hit what I feel, and secondly because the whole thing was an internet story from the beginning, which started with a discussion on de.sci.math. | 18,478 | 57,631 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 4, "equation": 65, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2024-26 | latest | en | 0.874649 |
https://vidyakul.com/cbse-class-12-maths-formulas-list-chapter-6-application-of-derivatives | 1,701,875,510,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100599.20/warc/CC-MAIN-20231206130723-20231206160723-00421.warc.gz | 664,233,757 | 32,981 | Class 12th Math Application Of Derivatives Formulas CBSE 2023
Chapter 6 Application of Derivatives Formulas
The NCERT notes for Class 12 Maths Chapter 6 Application of Derivatives answers all of the questions in the Class 12th Mathematics syllabus for 2022-23. These NCERT notes include a comprehensive set of questions and answers organised at an advanced level of difficulty, giving students ample opportunity to apply their knowledge.
Vidyakul intends to help students understand the basics of the chapter in a fun learning way. Students can easily track their progress on Vidyakul and work accordingly. Read on to know more.
Points to Remember
Some of the important points to remember from Class 12 Maths Chapter 6 are as follows:
• Application of Derivatives is widely used in engineering, science, economics, and social science, among other fields.
• One of the most important applications of derivatives is to find the best solution to problems. For example, finding the equations of tangent and normal to a curve at a point, or turning points on a function’s graph, will help us locate points where the largest or smallest value (locally) of a function occurs.
• Derivatives can also be used to calculate the intervals at which a function increases or decreases. Finally, it is useful to determine the approximate value of specific quantities.
• The derivatives are used to compute the rate of change, intervals of increasing or decreasing functions, the point at which the tangent is parallel or perpendicular, the approximate value of numbers, and so on.
• The first derivative test is used to find the extremum point of functions by analysing their first derivatives.
• The Second Order Derivative is the derivative of the given function’s first derivative.
• A derivative test in calculus uses a function’s derivatives to locate the critical points of a function and determine whether each point is a local maximum, a local minimum, or a saddle point. Derivative tests can also provide information about a function’s concavity.
Topics and Sub-topics
In this chapter, students will learn how to determine the rate of change of quantities, how to find turning points on the graph of a function, how to find the equations of tangent and normal to a curve at a point, and much more. The NCERT notes for 12th Class Maths Chapter 6 Applications of Derivatives have been designed by the top and experienced teachers at Vidyakul. These notes will help students tackle the exam questions with the correct approach.
Below we have provided the list of important topics explained by Vidyakul:
Topic Name Topic Name Rate of Change of Quantities Approximations Increasing and Decreasing Functions Maxima and Minima Tangents and Normals Maximum and Minimum Values of a Function in a Closed Interval
Download the FREE PDF of Application of Derivatives Class 12 Notes and start your preparation with Vidyakul! | 595 | 2,922 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2023-50 | latest | en | 0.914047 |
https://number.academy/111798 | 1,656,860,461,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104244535.68/warc/CC-MAIN-20220703134535-20220703164535-00701.warc.gz | 466,795,539 | 12,297 | # Number 111798
Number 111,798 spell 🔊, write in words: one hundred and eleven thousand, seven hundred and ninety-eight . Ordinal number 111798th is said 🔊 and write: one hundred and eleven thousand, seven hundred and ninety-eighth. Color #111798. The meaning of number 111798 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 111798. What is 111798 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 111798.
## What is 111,798 in other units
The decimal (Arabic) number 111798 converted to a Roman number is (C)(X)MDCCXCVIII. Roman and decimal number conversions.
#### Weight conversion
111798 kilograms (kg) = 246469.9 pounds (lbs)
111798 pounds (lbs) = 50711.2 kilograms (kg)
#### Length conversion
111798 kilometers (km) equals to 69469 miles (mi).
111798 miles (mi) equals to 179922 kilometers (km).
111798 meters (m) equals to 366787 feet (ft).
111798 feet (ft) equals 34077 meters (m).
111798 centimeters (cm) equals to 44015.0 inches (in).
111798 inches (in) equals to 283966.9 centimeters (cm).
#### Temperature conversion
111798° Fahrenheit (°F) equals to 62092.2° Celsius (°C)
111798° Celsius (°C) equals to 201268.4° Fahrenheit (°F)
#### Time conversion
(hours, minutes, seconds, days, weeks)
111798 seconds equals to 1 day, 7 hours, 3 minutes, 18 seconds
111798 minutes equals to 2 months, 3 weeks, 15 hours, 18 minutes
### Codes and images of the number 111798
Number 111798 morse code: .---- .---- .---- --... ----. ---..
Sign language for number 111798:
Number 111798 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
## Mathematics of no. 111798
### Multiplications
#### Multiplication table of 111798
111798 multiplied by two equals 223596 (111798 x 2 = 223596).
111798 multiplied by three equals 335394 (111798 x 3 = 335394).
111798 multiplied by four equals 447192 (111798 x 4 = 447192).
111798 multiplied by five equals 558990 (111798 x 5 = 558990).
111798 multiplied by six equals 670788 (111798 x 6 = 670788).
111798 multiplied by seven equals 782586 (111798 x 7 = 782586).
111798 multiplied by eight equals 894384 (111798 x 8 = 894384).
111798 multiplied by nine equals 1006182 (111798 x 9 = 1006182).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 111798
Half of 111798 is 55899 (111798 / 2 = 55899).
One third of 111798 is 37266 (111798 / 3 = 37266).
One quarter of 111798 is 27949,5 (111798 / 4 = 27949,5 = 27949 1/2).
One fifth of 111798 is 22359,6 (111798 / 5 = 22359,6 = 22359 3/5).
One sixth of 111798 is 18633 (111798 / 6 = 18633).
One seventh of 111798 is 15971,1429 (111798 / 7 = 15971,1429 = 15971 1/7).
One eighth of 111798 is 13974,75 (111798 / 8 = 13974,75 = 13974 3/4).
One ninth of 111798 is 12422 (111798 / 9 = 12422).
show fractions by 6, 7, 8, 9 ...
### Calculator
111798
#### Is Prime?
The number 111798 is not a prime number. The closest prime numbers are 111791, 111799.
#### Factorization and factors (dividers)
The prime factors of 111798 are 2 * 3 * 3 * 6211
The factors of 111798 are 1 , 2 , 3 , 6 , 9 , 18 , 6211 , 12422 , 18633 , 37266 , 55899 , 111798
Total factors 12.
Sum of factors 242268 (130470).
#### Powers
The second power of 1117982 is 12.498.792.804.
The third power of 1117983 is 1.397.340.037.901.592.
#### Roots
The square root √111798 is 334,362079.
The cube root of 3111798 is 48,173849.
#### Logarithms
The natural logarithm of No. ln 111798 = loge 111798 = 11,624449.
The logarithm to base 10 of No. log10 111798 = 5,048434.
The Napierian logarithm of No. log1/e 111798 = -11,624449.
### Trigonometric functions
The cosine of 111798 is 0,283045.
The sine of 111798 is 0,959107.
The tangent of 111798 is 3,388537.
### Properties of the number 111798
Is a Friedman number: No
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
## Number 111798 in Computer Science
Code typeCode value
PIN 111798 It's recommendable to use 111798 as a password or PIN.
111798 Number of bytes109.2KB
CSS Color
#111798 hexadecimal to red, green and blue (RGB) (17, 23, 152)
Unix timeUnix time 111798 is equal to Friday Jan. 2, 1970, 7:03:18 a.m. GMT
IPv4, IPv6Number 111798 internet address in dotted format v4 0.1.180.182, v6 ::1:b4b6
111798 Decimal = 11011010010110110 Binary
111798 Decimal = 12200100200 Ternary
111798 Decimal = 332266 Octal
111798 Decimal = 1B4B6 Hexadecimal (0x1b4b6 hex)
111798 BASE64MTExNzk4
111798 MD5a4febe0e3dfaa2c6f91bf85b8487ce74
111798 SHA1be09e2cb3ef8c7e183e0ac351975183323b56f7c
111798 SHA22402638840b0b04be197c98cfe3133ba514b6ed89611ca1a8e30df0f3b
111798 SHA256f1a76fd36a7792b1d9a407b614377652dbab64d2db49f9f35bff317a35c0422e
More SHA codes related to the number 111798 ...
If you know something interesting about the 111798 number that you did not find on this page, do not hesitate to write us here.
## Numerology 111798
### Character frequency in number 111798
Character (importance) frequency for numerology.
Character: Frequency: 1 3 7 1 9 1 8 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 111798, the numbers 1+1+1+7+9+8 = 2+7 = 9 are added and the meaning of the number 9 is sought.
## Interesting facts about the number 111798
### Asteroids
• (111798) 2002 CW242 is asteroid number 111798. It was discovered by LINEAR, Lincoln Near-Earth Asteroid Research from Lincoln Laboratory, Socorro on 2/11/2002.
## Number 111,798 in other languages
How to say or write the number one hundred and eleven thousand, seven hundred and ninety-eight in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 111.798) ciento once mil setecientos noventa y ocho German: 🔊 (Anzahl 111.798) einhundertelftausendsiebenhundertachtundneunzig French: 🔊 (nombre 111 798) cent onze mille sept cent quatre-vingt-dix-huit Portuguese: 🔊 (número 111 798) cento e onze mil, setecentos e noventa e oito Chinese: 🔊 (数 111 798) 十一万一千七百九十八 Arabian: 🔊 (عدد 111,798) مائة و أحد عشر ألفاً و سبعمائةثمانية و تسعون Czech: 🔊 (číslo 111 798) sto jedenáct tisíc sedmset devadesát osm Korean: 🔊 (번호 111,798) 십일만 천칠백구십팔 Danish: 🔊 (nummer 111 798) ethundrede og ellevetusindsyvhundrede og otteoghalvfems Dutch: 🔊 (nummer 111 798) honderdelfduizendzevenhonderdachtennegentig Japanese: 🔊 (数 111,798) 十一万千七百九十八 Indonesian: 🔊 (jumlah 111.798) seratus sebelas ribu tujuh ratus sembilan puluh delapan Italian: 🔊 (numero 111 798) centoundicimilasettecentonovantotto Norwegian: 🔊 (nummer 111 798) en hundre og elleve tusen, syv hundre og nitti-åtte Polish: 🔊 (liczba 111 798) sto jedenaście tysięcy siedemset dziewięćdzisiąt osiem Russian: 🔊 (номер 111 798) сто одиннадцать тысяч семьсот девяносто восемь Turkish: 🔊 (numara 111,798) yüzonbinyediyüzdoksansekiz Thai: 🔊 (จำนวน 111 798) หนึ่งแสนหนึ่งหมื่นหนึ่งพันเจ็ดร้อยเก้าสิบแปด Ukrainian: 🔊 (номер 111 798) сто одинадцять тисяч сiмсот дев'яносто вiсiм Vietnamese: 🔊 (con số 111.798) một trăm mười một nghìn bảy trăm chín mươi tám Other languages ...
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## Comment
If you know something interesting about the number 111798 or any natural number (positive integer) please write us here or on facebook. | 2,526 | 7,497 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2022-27 | latest | en | 0.632238 |
https://www.i2tutorials.com/machine-learning-tutorial/decision-trees/ | 1,632,385,879,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057417.92/warc/CC-MAIN-20210923074537-20210923104537-00236.warc.gz | 853,022,360 | 15,735 | / / Decision Trees
# Decision Trees
Decision tree is a decision tool that uses a tree-like graph to represent their possible consequences or outcomes, including chance event outcomes, resource costs, and effectiveness. It is a like flowchart structure in which each internal node represents a test on an attribute, each branch represents the outcome of the test, and each leaf node represents a decision taken after computing all attributes.
Decision Tree algorithms are considered to be one of the best and mostly used supervised learning methods. Tree type of methods empower predictive models with high accuracy, stability and easy understanding. Decision trees can also map non-linear relationships quite well. They can handle both regression and classification problems. Hence, they are referred as CART (Classification and Regression Trees).
### Notation used in Decision trees:
1. Root Node: It represents whole population or sample and further gets divided into more homogeneous sets.
2. Splitting: Procedure of dividing a node into two or more sub-nodes.
3. Decision Node: When a node splits further into sub-nodes, then the node is called decision node.
4. Leaf/ Terminal Node: Nodes which cannot be split further is called Leaf or Terminal node.
5. Pruning: Process of removing sub-nodes of a decision node, is called pruning. To make it opposite process of splitting.
6. Branch / Sub-Tree: A part of entire tree is called branch or sub-tree.
7. Parent and Child Node: A node, which is split into sub-nodes is called parent whereas sub-nodes are the child of parent node.
### Types of Decision Trees
Decision Trees are classified based on type of Target variables. It is classified into two types:
1. Categorical Variable Decision Tree: Decision Tree which has target as categorical variable is called a Categorical variable decision tree.
2. Continuous Variable Decision Tree: Decision Tree which has target as continuous variable then it is called Continuous Variable Decision Tree.
### Attribute Selection Measures
If the dataset consists of N attributes or features then deciding which attribute to put at the basis and at nodes. By just randomly selecting any node to be the basis can’t solve the difficulty. By following random approach, it’s going to give us bad results with low accuracy or performance.
For selecting attributes, we consider following measures:
Entropy
Information gain
Gini index
Gain Ratio
Reduction in Variance
Chi-Square
These criteria will calculate values for each attribute. The values are sorted, and attributes are placed within the tree by following the order i.e., the attribute with a high value is placed at the basis.
While using Information Gain as a main point, we assume attributes to be categorical, and for the Gini index, attributes are assumed to be continuous.
### Entropy
Entropy may be a measure of the randomness within the information being processed. the upper the entropy, the harder it’s to draw any conclusions from that information. Flipping a coin is an example of an action that gives information that’s random.
The Entropy will be maximum when the probability is 0.5 because it displays perfect randomness in the data and there is no chance if perfectly defining the outcome.
A node with zero entropy is known as leaf node.
Mathematically Entropy for 1 attribute is represented as:
Where S → Current state, and Pi → Probability of an event i of state S or Percentage of class i in a node of state S.
Mathematically Entropy for multiple attributes is represented as:
where T→ Current state and X → Selected attribute
### Information Gain
Information gain or IG is a statistical property which measures how better the given attribute differentiate the training examples according to their target classification. Building a decision tree is all about finding an attribute that gives the outcome as highest information gain and the smallest entropy.
Information gain is a reduce in entropy. It calculates the difference between entropy before split and average entropy after split of the dataset based on given attribute values.
Mathematically, IG is represented as:
In a much simpler way, we can conclude that:
Where before is the dataset before the split, K is the number of subsets generated by the split, and j after is subset j after the split.
### Gini Index
Gini index is a type of cost function which is used to evaluate splits in the dataset. It is computed by subtracting the sum of the squared probabilities of each class from one. It supports larger partitions and easy to implement whereas information gain favors smaller partitions with distinct values.
Gini Index deals with the categorical target variable Success or Failure. It performs only Binary separations.
Gini index is proportional to Homogeneity, higher the value of Gini index higher the homogeneity.
#### Steps to Calculate Gini index
1. Compute Gini for sub-nodes, using the formula for success(p) and failure(q) (p²+q²).
2. Calculate the Gini index for split using the weighted Gini score of each split node.
3. (CART) Classification and Regression Tree uses the Gini index method to create split points.
### Gain ratio
Information gain is biased towards choosing attributes with a large number of values as root nodes which means it only prefers the attribute with a large number of distinct values.
Gain ratio overcomes the limitations of information gain by taking into account the number of branches that would result before making the split. It corrects information gain by taking the intrinsic information of a split into consideration.
Where before is the dataset before the split, K is the number of subsets generated by the split, and j after is subset j after the split.
### Reduction in Variance
Reduction in variance is an algorithm which can be used for continuous target variables that is regression problems. This algorithm uses the general formula of variance to choose the best split. The split which has lower variance is selected as the criteria to split the population:
Where X-bar is the mean of the values, X is actual and n is the number of values.
### Chi-Square
CHAID abbreviated as Chi-squared Automatic Interaction Detector. It finds out the statistical significance of the differences between child nodes and parent node. We can measure it by the sum of squares of standardized differences between observed and expected frequencies of the target variable.
It deals with the categorical target variable Success or Failure. It can able to perform two or more splits. Higher the Chi-Square value, higher the statistical significance of differences between child node and Parent node.
Mathematically, Chi-squared is represented as:
#### Steps to Calculate Chi-square for a split:
1. Compute Chi-square for an individual node by calculating the deviation for Success and Failure.
2. Calculated Chi-square of Split using Sum of all Chi-square of success and Failure of each split node.
1. Easy to Understand: Decision tree output is very easy to understand. It may not require any statistical knowledge to read and analyze them. Its graphical representation is very spontaneous and users can easily relate their hypothesis.
2. Useful in Data exploration: Decision tree is one of the fastest ways to identify most significant variables and relation between them. With the help of decision trees, we can create new variables / features that has better power to predict target variable. It is also used in data exploration stage.
3. Decision trees implicitly perform variable broadcast or feature selection.
4. Decision trees need relatively little effort from users for data preparation.
5. Less data cleaning required: It does not require more data cleaning as some other modeling techniques. It cannot be influenced by outliers and missing values to a fair degree.
6. Data type is not a constraint: It can handle both numerical and categorical variables.
7. Non-Parametric Method: Decision tree is considered to be a non-parametric method which means that decision trees have no assumptions about the space distribution and the classifier structure.
8. Non-linear relationships between parameters do not affect tree performance.
9. The number of hyper-parameters to be tuned is almost null. | 1,630 | 8,314 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2021-39 | latest | en | 0.926234 |
https://lessonplanet.com/search?keywords=esl+comparative+adjectives | 1,624,223,763,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488257796.77/warc/CC-MAIN-20210620205203-20210620235203-00213.warc.gz | 336,382,660 | 29,693 | We found38 reviewed resources for esl comparative adjectives
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In this ESL instructional activity, students explain 15 words to a partner. Students use language like, "It's as big as.." and "I think it's more dangerous than..."
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Best Friend - Friend from Hell
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Students practice in a number of areas: expressing opinions, using comparatives and superlatives, descriptive adjectives and reported speech. The overall concept of this lesson can be applied to many other topics of discussion.
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In this adjectives ESL worksheet, students analyze 7 pictures that compare 2 different objects. One of the objects has an arrow pointing to it. Students match the picture with the adjective that best describes the indicated object.
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Lesson Planet | 1,275 | 5,994 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2021-25 | latest | en | 0.897832 |
https://www.twblogs.net/a/5f05232254de47abadd9c142 | 1,675,093,441,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499819.32/warc/CC-MAIN-20230130133622-20230130163622-00822.warc.gz | 1,031,185,413 | 10,378 | # LeetCode718. 最長重複子數組(python)
### 1. 問題
A: [1,2,3,2,1]
B: [3,2,1,4,7]
1 <= len(A), len(B) <= 1000
0 <= A[i], B[i] < 100
### 2. 解法
class Solution:
def findLength(self, A: List[int], B: List[int]) -> int:
#dp[i][j] = longest(A[:i],B[:j])
res = -float('inf')
dp = [[0 for j in range(len(B)+1)] for i in range(len(A)+1)]
for i in range(1,len(A)+1):
for j in range(1,len(B)+1):
if A[i-1] == B[j-1]:
dp[i][j] = dp[i-1][j-1] + 1
else:
dp[i][j] = 0
res = max(dp[i][j],res)
return res | 223 | 483 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 2, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2023-06 | longest | en | 0.323797 |
https://electronics.stackexchange.com/questions/175179/automotive-power-supply-considerations-for-selecting-fuse | 1,716,028,176,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057379.11/warc/CC-MAIN-20240518085641-20240518115641-00271.warc.gz | 201,738,717 | 41,366 | # Automotive power supply - considerations for selecting fuse
A power supply circuit for a device that runs off the 12V battery of a car, is shown below.
simulate this circuit – Schematic created using CircuitLab
LM7805 is used to generate a 5V supply for the circuit - max load current is 200mA. A forward diode and a TVS diode are used for negative/positive spike protection (load dump). And a 10R, 2W resistor is used to limit the inrush current due to the 470uF cap at the input of LM7805. To prevent tripping during load dump or inrush current events, I am using a 3A slow-blow fuse (1.27 A2s).
Questions:
1. One normal current overload situation is inrush current due to the 470uF cap (inrush current = 1.2A but might be 4x that if a huge positive spike occurs on start up). How do I calculate the inrush current pulse duration?
2. Another normal current overload situation is load dump wherein, as per ISO-7637-2 pulse #5, the pulse duration can be as much as ~500ms (Vp ~ 100V). Vc for the TVS diode I am using is 30V so the circuit could potentially see a current pulse of (100-30)/10 = 7A for 500ms due to load dump. Is my calculation correct?
3.Considering the above, I am using a 3A slow-blow fuse but this might actually be useless considering that LM7805 should prevent more than 1.5A being drawn through the circuit and if LM7805 input node gets shorted, then a maximum of 12V/10R = 1.2A (nominal) would flow through the circuit. This 1.2A would be sufficient to burn out resistor & is therefore a concern. However, if I use a lower current rating for the fuse (say 1 or 1.5A), it might blow up during either of the 2 normal current overload situations above. Plus if I take derating factor of 0.5x due to thermal cycling and temperature, a 1 or 1.5A fuse might blow that much more easily couple of months down the line. Also, I don't think I can increase the resistor value much - the max voltage drop I can afford across it is ~3V (min LM7805 i/p voltage = 7V and there will be voltage drop across the FWD diode and fuse in addition to R) which for a 200mA peak load current, means 15 ohm max.What would be the optimal tradeoff here with respect to selecting the fuse?
1. What sort of situations could actually result in a short to GND at any of the nodes connected to 7805 i/p? i.e., assuming the device is mechanically robust, should I still be concerned about a potential short to GND there?
• Where did the circuit come from i.e. was the circuit from a bona fide auto site that offers good ideas for this sort of thing? Jun 12, 2015 at 14:42
• The circuit is my own - but is based on research and examples. Specifically, I believe the general structure of the circuit might be common (for a linear regulator circuit- there are also switched mode circuits). But the values would differ depending on the particular application (mainly load current, I would think). Jun 12, 2015 at 15:25
Firstly think about what the purpose of the fuse is. Basically the thing you don't want is for whatever is being protected by the fuse - components, wiring, connectors - to go up in flames before the fuse breaks under fault conditions (either single or double depending on how safe you want your design to be). Remember that when the fault occurs the system is already damaged and the fuse is only there to prevent a dangerous failure.
So lets analyse the failure paths in your circuit:
1. the wiring or connectors between the fuse and the unit shorting out.
2. D1 failing open or closed circuit.
3. 10ohm resistor failing open or closed circuit.
4. D2, capacitor or regulator failing open or closed circuit.
In scenario (1) the fuse just needs to blow before the wiring does, so unless the wires are particularly thin even a very high value would be fine. In (2) and (3) there is no short circuit risk. Scenario (4) is then where you start to see a problem. If any of these fail closed circuit then current will flow in a loop through D1 and the 10ohm resistor. This will put more than 10W through your resistor which exceeds its rating. Even so it may not pose a dangerous risk (the resistor will likely just burn out and either short - which will then blow the fuse, or open circuit), but you might want to either reduce the fuse to a size that protects for this or get a non-flammable fusing resistor (which you probably need for the load dump rating anyway). As you identified you may not be able to reduce the fuse enough due to the inrush/load dump current anyway.
As an aside, note that the battery voltage can easily fall to 7V or less during cold cranking conditions. Your circuit as it stands won't provide enough margin to prevent the regulator dropping out so you might need to reconsider your design if this is a problem.
Also keep in mind that most modern vehicles have a central load dump surge suppressor on the alternator so individual modules don't need to protect against this. This makes achieving cold cranking voltage operation much simpler.
• Thanks. This would actually be going in some old models so I might not be able to bank on the central load dump surge suppressor. Jun 12, 2015 at 15:20
• Yes, that will be a problem with older cars. Have you thought about using an automotive grade regulator? Maxim, infineon etc make ones with built in load dump, reverse polarity protection etc. Automotive supplies for older cars need to be pretty tough and it might be worth the extra cost to use a specialised device.
– Jon
Jun 12, 2015 at 15:30
• Agreed and that is what I will probably insist on using the next time! Just that for this particular product, cost was a big factor and I needed to design something robust within a budget. The PCB is already out and field tested - works in nominal conditions. But want to make sure I am not missing anything - particularly with regards to the fuse because it seems a little bit tricky to get it right. Jun 12, 2015 at 16:18
• In that case I would just go for a 2-3A slow blow as you first thought. This will protect you from poor wiring by an installer (which in my experience is the most common fault) and limits the power that can go into the system at around 30W, which isn't going to be able to do much damage.
– Jon
Jun 12, 2015 at 17:19 | 1,494 | 6,232 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2024-22 | latest | en | 0.895392 |
http://oeis.org/A209666 | 1,548,118,723,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583822341.72/warc/CC-MAIN-20190121233709-20190122015709-00564.warc.gz | 164,835,175 | 4,024 | This site is supported by donations to The OEIS Foundation.
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A209666 T(n,k) = count of degree k monomials in the complete homogeneous symmetric polynomials h(mu,k) summed over all partitions mu of n. 4
1, 2, 7, 3, 18, 55, 5, 50, 216, 631, 7, 118, 729, 2780, 8001, 11, 301, 2621, 12954, 45865, 130453, 15, 684, 8535, 55196, 241870, 820554, 2323483, 22, 1621, 28689, 241634, 1307055, 5280204, 17353028, 48916087, 30, 3620, 91749, 1012196, 6783210, 32711022, 124991685, 401709720, 1129559068 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 1,2 LINKS Alois P. Heinz, Rows n = 1..141, flattened Wikipedia, Symmetric Polynomials EXAMPLE Table starts as: 1; 2, 7; 3, 18, 55; 5, 50, 216, 631; 7, 118, 729, 2780, 8001; MAPLE b:= proc(n, i, k) option remember; `if`(n=0, 1, `if`(i<1, 0, add(b(n-i*j, i-1, k)*binomial(i+k-1, k-1)^j, j=0..n/i))) end: T:= (n, k)-> b(n\$2, k): seq(seq(T(n, k), k=1..n), n=1..10); # Alois P. Heinz, Mar 04 2016 MATHEMATICA h[n_, v_] := Tr@ Apply[Times, Table[Subscript[x, j], {j, v}]^# & /@ Compositions[n, v], {1}]; h[par_?PartitionQ, v_] := Times @@ (h[#, v] & /@ par); Table[Tr[(h[#, k] & /@ Partitions[l]) /. Subscript[x, _] -> 1], {l, 10}, {k, l}] CROSSREFS Main diagonal is A209668; row sums are A209667. Sequence in context: A076992 A138751 A112303 * A089124 A210662 A229610 Adjacent sequences: A209663 A209664 A209665 * A209667 A209668 A209669 KEYWORD nonn,tabl AUTHOR Wouter Meeussen, Mar 11 2012 STATUS approved
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Last modified January 21 19:57 EST 2019. Contains 319350 sequences. (Running on oeis4.) | 747 | 1,900 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2019-04 | latest | en | 0.45126 |
http://betterlesson.com/lesson/reflection/3646/what-do-i-do-with-that-x | 1,477,575,085,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988721278.88/warc/CC-MAIN-20161020183841-00270-ip-10-171-6-4.ec2.internal.warc.gz | 27,091,440 | 20,900 | Reflection: Grappling with Complexity Simplifying Linear Expressions by Combining Like Terms - Section 2: Guided Problem Solving
One thing I overlooked was that many students would not know how to handle a variable without a visible coefficient. So when encountering 3x + x they would not know what to do. It is a simple fix, but I think I should have made sure to mention that x means 1x. Once this was explained, students were able to simplify with it. To be even more clear I could remind students that 3x = x + x + x, therefore 3x + x = (x + x + x) + x = 4x.
What do I do with that x?
Grappling with Complexity: What do I do with that x?
Simplifying Linear Expressions by Combining Like Terms
Unit 4: Expressions and Equations
Lesson 1 of 20
Big Idea: Students add 5 apples and 3 bananas to 6 apples and 2 bananas algebraically.
Print Lesson
27 teachers like this lesson
Standards:
Subject(s):
Math, Expressions (Algebra), combining like terms, simplifying expressions, equation
50 minutes
Grant Harris
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Environment: Suburban | 439 | 1,773 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2016-44 | longest | en | 0.924153 |
https://jonathan-ck-chao.medium.com/the-blind-75-leetcode-series-construct-binary-tree-from-preorder-and-inorder-traversal-84ee0621b563 | 1,669,745,436,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710710.91/warc/CC-MAIN-20221129164449-20221129194449-00558.warc.gz | 373,121,379 | 37,124 | # The Blind 75 Leetcode Series: Construct Binary Tree from Preorder and Inorder Traversal
Today, we are working on 105. Construct Binary Tree from Preorder and Inorder Traversal
Given two integer arrays `preorder` and `inorder` where `preorder` is the preorder traversal of a binary tree and `inorder` is the inorder traversal of the same tree, construct and return the binary tree.
We are given 2 lists, one as the preorder traversal result of a tree and another as the inorder traversal result of the same tree. From these lists, we want to figure out what this tree looks like.
1. do we expect empty tree? (empty list)
2. are the values in the lists unique?
If the interviewers tell us that the tree always contains some nodes and the values are all unique, then we can start.
If you are not familiar with the pre-order and in-order traversals, basically pre-order traversal records the value first, then go left, then go right. It repeats the process at every node until we go through the entire tree. In-order traversal goes left first until it can’t. Then it records the value. Then it goes right. Once it goes to the next node, the process repeats.
Knowing these properties, we can say a few things:
1. the first value on the pre-order list is the root since pre-order traversal would record the value of the node first.
2. Knowing the value of root, since all values are unique, we can find the index of the value in the in-order traversal. Once we do, every value on the left of the root value is on the left side of the tree, and every value on the right belongs to the right side of the tree.
3. Now that we know the size of the tree on the left and on the right, we go back to pre-order list to slice the list. Since pre-order list would always have values from the left tree recorded first, if we know from the in-order list that there should be 3 nodes on the left tree, we can go ahead and slice the first 3 values from the pre-order list. These values should match. Try testing it.
4. We now have two sets of pre-order and in-order combos. We can feed it into the same function again to recursively build this tree.
To put everything into code, it looks like
`def buildTree(preorder, inorder): if not preorder or not inorder: return root = TreeNode(preorder[0]) # find the index in the in-order list idx = inorder.index(root.val) # Slice the two lists into left-half and right-half. Knowing how to slide the lists is the key to this problem left_preorder, right_preorder = preorder[1:idx+1], preorder[idx+1:] left_inorder, right_inorder = inorder[:idx], inorder[idx+1:] # recursively build the tree root.left = self.buildTree(left_preorder, left_inorder) root.right = self.buildTree(right_preorder, right_inorder) return root`
What’s the time complexity here? We are going over the list recursively due to that `.index` operation, which takes linear time. We end up with O(N²) overall.
`def buildTree(preorder, inorder): if not preorder or not inorder: return root = TreeNode(preorder[0]) # O(N) idx = inorder.index(root.val) left_preorder, right_preorder = preorder[1:idx+1], preorder[idx+1:] left_inorder, right_inorder = inorder[:idx], inorder[idx+1:] # recursion on O(N) makes it O(N^2) root.left = self.buildTree(left_preorder, left_inorder) root.right = self.buildTree(right_preorder, right_inorder) return root`
Can we do better? What if we can eliminate the `.index`? That seems like the biggest bottleneck. If we can reduce that to O(1), we can reduce the overall time complexity to O(N). What type of “tracking” can give us O(1) read time? That sounds like a job for dictionary (hash table).
We start with doing one iteration through the entire inorder list to get the positions of each number. Then we do the same operation, except this time we find the idx through `hashtable.get(val)` . But to do that, we can’t just slice the list since it’ll change the index position. We will need to find a way to correctly track the in-order list’s index.
Instead of slicing the in-order list, we can use 2 pointers, left and right, to determine how we want to slice it without actually slicing the list.
We still want to get the first element of the pre-order list, so we want to pass in the pre-order list as part of the recursive function inputs as well. Then we need to pass in the tracker to determine the position of the elements in in-order list.
So our recursive functions will take…
`def recursive(left, right, preorder, inorder_index_map)`
The logic is almost the same inside the recursive. We first take the first element of the pre-order list. Since we are not slicing the list this time, we want to `pop` it from the list. `pop(0)` is a O(N) operation, so we want to make sure to convert the pre-order list to a `deque`, so we can do a O(1) `popleft()`.
We now just need to determine the correct value for `left` and `right` for each recursion.
Going back to the property of in-order list, we know that the left side of the target idx belongs to the left tree and the right side belongs to the right tree. So the `left` and `right` for the left tree is simply the same `left` and `inorder_index_map[root_val] — 1`, where `inorder_index_map` is the tracker we built and `root_val` is the value for the current node.
Similarly, for the right tree, we have the `right` being the same and `left` becomes `inorder_index_map[root_val] + 1`.
We can now put everything together.
`from collections import dequedef recursive(left, right, preorder, inorder_index_map): # if there are no elements to construct the tree if left > right: return root_val = preorder.popleft() root = TreeNode(root_val) root.left = recursive(left, inorder_index_map[root_val] - 1, preorder, inorder_index_map) root.right = recursive(inorder_index_map[root_val] + 1, right, preorder, inorder_index_map)return rootdef buildTree(preorder, inorder]): # so we can popleft with O(1) preorder = deque(preorder) # tracker for the value: idx of in-order list inorder_index_map = {} for index, value in enumerate(inorder): inorder_index_map[value] = index return recursive(0, len(preorder) - 1, preorder, inorder_index_map)`
And now, because we eliminated the `.index` and we did not add any more O(N) operation, we are now only looking at O(N) overall time complexity.
`def recursive(left, right, preorder, inorder_index_map): if left > right: # O(1) return root_val = preorder.popleft() # O(1) root = TreeNode(root_val) # O(1) root.left = recursive(left, inorder_index_map[root_val] - 1, preorder, inorder_index_map) root.right = recursive(inorder_index_map[root_val] + 1, right, preorder, inorder_index_map)return rootdef buildTree(preorder, inorder]): preorder = deque(preorder) # building tracker: O(N) inorder_index_map = {} for index, value in enumerate(inorder): inorder_index_map[value] = index # overall recursive: O(N) return recursive(0, len(preorder) - 1, preorder, inorder_index_map)`
That’s it! Another problem down.
--
--
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## Jonathan Chao
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I am a software developer who has been in this industry for close to a decade. I share my experience to people who are or want to get into the industry | 1,811 | 7,565 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2022-49 | latest | en | 0.878592 |
https://technewspedia.com/divide-in-microsoft-excel-step-by-step-guide-%E2%96%B7-2020/ | 1,642,607,803,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301475.82/warc/CC-MAIN-20220119155216-20220119185216-00058.warc.gz | 603,892,848 | 15,911 | # Divide in Microsoft Excel Step by Step Guide â· 2020
Microsoft Excel is a tool with multiple functions to manage finances in an environment domestic or business. The software allows Solve mathematical problems quickly and easily as well how to split cell data automatically.
Additions, subtractions, multiplications and divisions are possible in Excel. No need to resort to a calculator. With only specify cells corresponding with a simple formula, the operation I know will resolve immediately.
You can make a division On any cell, row or column and the results they will be very successful. Do you want to know how to split cell data in Excel?. Keep reading, as we present you a step-by-step guide.
## What are the best formulas to divide anything in Excel?
All mathematical formula included in Excel should start with a sign of âEqualâ (=). On the other hand, the divisions are characterized by using the division operator (/). For example, a division into Excel is expressed as âsameâ, followed by the ânumeratorâ, accompanied by the âoperatorâ and culminates in the âdenominatorâ (= 20/2). To obtain the result, it is only necessary to press the key âEnterâ.
There are two methods to perform a division in Excel. First of all, you can enter the complete formula in any cell. Either by clicking and typing on the same grid, or use the formula bar on the ribbon. These are the best formulas to divide anything in Excel.
For example, with the first method Try introducing the following mathematical problem: = 20/2 By pressing the key âEnterâ, the result will appear above the cell that you have previously selected.
The second method, implies the possibility of perform the mathematical operation with a numerator and denominator located in different parts of the spreadsheet. They can be in a row, column, or multiple cells at once.
For example, letâs say the number 20 is in the cell B4, Meanwhile he number two (2) is in the cell G6. All you have to do is enter division formula on any other grid. click about C8, or about a cell of your preference.
Enter the sign âEqualâ (=) followed by the position of the numerator (B4), then the division operator (/) and culminates with the location of the denominator (G6), that is to say, (= B4 / G6) to perform the mathematical operation, proceed to press âEnterâ. The difference between both formulas is that, in the first method, the result happens to occupy the cell where the formula was entered, so the values ââdisappear. While the second method, preserves the numerator and denominator values.
Similarly, when it comes to split multiple figures into columns or rows for one between the same number, it is necessary to add the sign â\$â in the formula, to specify that the number is absolute for all cells. If the absolute number is in C2, you must divide it with the first value of the row or column. Thus, the formula is expressed like this: = B2 \$ C \$ 2.
## Steps to perform a division in Microsoft Excel automatically
Perform mathematical operations in excel itâs not complicated. The formulas are easy to learn and you only need to open one new spreadsheet.
Next, we are going to show you the steps to perform a division in Microsoft Excel automatically:
### Data
Dividing data in Excel is quite simple, you just have to enter a formula on any cell:
• Position yourself on a cell of your preference. For example B2.
• Letâs say you want to know how much 20 is between 2. To do this, proceed to enter the following formula: = 20/2.
• Press the key âEnterâ. The result will appear in the box B2.
### Cells
The process it is similar to the previous one. The difference is in the location of the values.
The steps to follow to divide into cells are the following:
• Enter the numerator in a box. For example, B2.
• Enter the denominator in a continuous box. For example, C2.
• In the next cell, D2, enter the division formula. Only instead of entering the numbers, you should add the locations, that is: = B2 / C2.
• Press âEnterâ on the keyboard to perform division. The result will appear in cell D2, without modifying the grids B2 and C2.
### Columns
In Excel it is also possible split multiple cells in a column to get a total result.
Starts with:
• Enter the entries in a column of your choice. For example, letâs say the numbers are contained from the cell B2 through B20, and you want split all grids by the same number located in C2.
• Click on any other cell. For example, D2. Now proceed to Enterokay the following formula: = B2 / \$ C \$ 2.
• The result will appear in cell D2. In the lower right corner of cell, you will find a small square.
• Click on the box and drag it to the box D20 for the formula to apply to the rest of the values ââof column B2 to B20.
### Rows
The operation it is executed in the same way as the previous procedure. Except that, instead of a column, it is done with a row.
Letâs see how to do it:
• Enter the entries in a row of your choice. For example, in the row 4 from the column B. (B4, C4, D4, E4, F4, G4)
• In row 5 (B5) enter the number by which the entries should be divided.
• Click on the next row (B6) and proceed to enter the formula: = B4 / \$ B \$ 5
• By pressing âEnterâ, the result will appear in cell B6.
• Now, select the grid again B6 and press the lower right corner, on a black square.
• Drag the square to the right, up to box G6. In this way, all values ââwill be divided between value of B5.
If you have any questions, leave them in the comments, we will answer you as soon as possible, and it will also be of great help to more members of the community. Thank you! đ
Rate this post | 1,396 | 5,715 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2022-05 | latest | en | 0.917545 |
http://www.mathworks.com/matlabcentral/cody/problems/2264-gjam-2011-africa-qualifier-c-house-maximum-area-rectangle | 1,477,499,532,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988720962.53/warc/CC-MAIN-20161020183840-00008-ip-10-171-6-4.ec2.internal.warc.gz | 549,337,575 | 12,951 | Cody
# Problem 2264. GJam 2011 Africa Qualifier C: House, maximum area rectangle
This Challenge is derived from GJam 2011 Africa: Building a House. Google Code Jam 2014 Kicks Off its Qualifier round April 11. GJam Registration. The Test Suite, at the bottom, contains a full GJam file input read (fgetl, string), process, and Output example.
Determine maximum area rectangular House that can be placed on a plot given a text terrain map of G-Grass,S-Shrubs,R-Rock,W-Water, and T-Tree. The house can not be placed over Water, Trees, or Rocks. Square is a special rectangle.
Input: M, Text Array of GWRST
Output: A, maximum House area that can be placed
Example:
```GGTGG A=9
TGGGG
GSSGT
GGGGT
GWGGG
RGTRT
RTGWT
WTWGR
```
Additional GJam solutions can be found at Example GJam Matlab solutions. Select Find Solutions, change Language to Matlab. There were only 50 Qualifier Matlab entrants in 2013 and a mere 2 Matlab users achieved round 2.
### Solution Stats
70.0% Correct | 30.0% Incorrect
Last solution submitted on Jan 24, 2015
Tags
MATLAB and Simulink resources for Arduino, LEGO, and Raspberry Pi
Apply Today | 313 | 1,127 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2016-44 | longest | en | 0.800143 |
https://ceres-solver.googlesource.com/ceres-solver/+/84436f791c2f76a82f56190ba9d744c904c07551/examples/sampled_function/ | 1,718,379,569,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861567.95/warc/CC-MAIN-20240614141929-20240614171929-00762.warc.gz | 136,759,177 | 2,898 | tree: 6b6f199e968f8c28aaa2637530f7f0cbc5bc4841 [path history] [tgz]
Sampled Functions
It is common to not have an analytical representation of the optimization problem but rather a table of values at specific inputs. This commonly occurs when working with images or when the functions in the problem are expensive to evaluate. To use this data in an optimization problem we can use interpolation to evaluate the function and derivatives at intermediate input values.
There are many libraries that implement a variety of interpolation schemes, but it is difficult to use them in Ceres' automatic differentiation framework. Instead, Ceres provides the ability to interpolate one and two dimensional data.
The one dimensional interpolation is based on the Cubic Hermite Spline. This interpolation method requires knowledge of the function derivatives at the control points, however we only know the function values. Consequently, we will use the data to estimate derivatives at the control points. The choice of how to compute the derivatives is not unique and Ceres uses the Catmull-Rom Spline variant which uses `0.5 * (p_{k+1} - p_{k-1})` as the derivative for control point `p_k.` This produces a first order differentiable interpolating function. The two dimensional interpolation scheme is a generalization of the one dimensional scheme where the interpolating function is assumed to be separable in the two dimensions.
This example shows how to use interpolation schemes within the Ceres automatic differentiation framework. This is a one dimensional example and the objective function is to minimize `0.5 * f(x)^2` where `f(x) = (x - 4.5)^2`.
It is also possible to use analytical derivatives with the provided interpolation schemes by using a `SizedCostFunction` and defining the ``Evaluate` function. For this example, the evaluate function would be:
```bool Evaluate(double const* const* parameters, double* residuals, double** jacobians) const {
if (jacobians == nullptr || jacobians[0] == nullptr)
interpolator_.Evaluate(parameters[0][0], residuals);
else
interpolator_.Evaluate(parameters[0][0], residuals, jacobians[0]);
return true;
}
``` | 467 | 2,159 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2024-26 | latest | en | 0.836476 |
https://socratic.org/questions/5a325e7a11ef6b6a98c43877 | 1,624,518,061,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488551052.94/warc/CC-MAIN-20210624045834-20210624075834-00453.warc.gz | 480,858,481 | 6,006 | # Question #43877
##### 1 Answer
Feb 25, 2018
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\cos \left(x + y\right) - \cos \left(y\right)}{- x \sin \left(y\right) - \cos \left(x + y\right)}$
#### Explanation:
Differentiate both sides with respect to $x$. This means that every time we differentiate a term containing $y$, we should end up with an instance of $\frac{\mathrm{dy}}{\mathrm{dx}}$.
$\frac{d}{\mathrm{dx}} \left(x \cos \left(y\right)\right) = \frac{d}{\mathrm{dx}} \left(\sin \left(x + y\right)\right)$
$\cos \left(y\right) \frac{d}{\mathrm{dx}} \left(x\right) + x \frac{d}{\mathrm{dx}} \cos \left(y\right) = \cos \left(x + y\right) \frac{d}{\mathrm{dx}} \left(x + y\right)$
$- x \sin \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}} + \cos \left(y\right) = \left(1 + \frac{\mathrm{dy}}{\mathrm{dx}}\right) \cos \left(x + y\right)$
Solve for $\frac{\mathrm{dy}}{\mathrm{dx}} :$
$- x \sin \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}} + \cos \left(y\right) = \cos \left(x + y\right) + \frac{\mathrm{dy}}{\mathrm{dx}} \cos \left(x + y\right)$ (Multiply out on the right side)
$- x \sin \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{\mathrm{dy}}{\mathrm{dx}} \cos \left(x + y\right) = \cos \left(x + y\right) - \cos \left(y\right)$ (Isolate all terms containing $\frac{\mathrm{dy}}{\mathrm{dx}}$ on the left side, move all other terms to the right)
$\frac{\mathrm{dy}}{\mathrm{dx}} \left(- x \sin \left(y\right) - \cos \left(x + y\right)\right) = \cos \left(x + y\right) - \cos \left(y\right)$ (Factor out $\frac{\mathrm{dy}}{\mathrm{dx}}$)
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\cos \left(x + y\right) - \cos \left(y\right)}{- x \sin \left(y\right) - \cos \left(x + y\right)}$
(Divide both sides by $- x \sin \left(y\right) - \cos \left(x + y\right)$) | 659 | 1,778 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 15, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2021-25 | latest | en | 0.354387 |
https://www.dataunitconverter.com/kibibit-per-second-to-byte-per-second | 1,686,380,539,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224657144.94/warc/CC-MAIN-20230610062920-20230610092920-00458.warc.gz | 812,360,706 | 13,017 | # Kibps to Bps Converter - CONVERT Kibibits Per Second to Bytes Per Second
Copy Link & Share
Input Kibibits Per Second - and press Enter
Kibps
Sec
Min
Hr
Day
Sec
Min
Hr
Day
Quickly and accurately convert between Kibibits Per Second and Bytes Per Second with our free online tool. Learn about the conversion formula and calculation steps. Get precise results and save time with DataUnitConverter.
## Recent Conversions
History Empty ! No Recent Conversions.
## How to use Kibibits Per Second to Bytes Per Second Converter
Kibps to Bps Calculator Tool convert the data transfer rate from Kibibits Per Second to Bytes Per Second.
It is very easy to use, just follow the below steps.
• Type the value in Kibps input box and click CONVERT button or simply hit ENTER key.
• The calculator will process the conversion with the highest accuracy and display the result.
• Use the Copy button to copy the result to clipboard.
• Click on the Swap⇄ button to reverse the conversion direction.
You can also change the source and target units in the drop-downs and quickly navigate to an entirely different conversion. Alternatively, switch to Data Size Converter for calculating the data storage size.
If you are looking to convert from one number system to another, such as binary, decimal, octal, or hexadecimal, try out the Number Base Converters.
## Kibps to Bps Formula and Manual Conversion Steps
Kibibit and Byte are units of digital information used to measure storage capacity and data transfer rate. Kibibit is a binary unit where as Byte is one of the very basic digital unit. One Kibibit is equal to 1024 bits. One Byte is equal to 8 bits. There are 0.0078125 Kibibits in one Byte. - view the difference between both units
Source Data UnitTarget Data Unit
Kibibit (Kibit)
Equal to 1024 bits
(Binary Unit)
Byte (B)
Equal to 8 bits
(Basic Unit)
The formula of converting the Kibibits Per Second to Bytes Per Second is represented as follows :
Bps = Kibps x 1024 / 8
Now let us apply the above formula and see how to manually convert Kibibits Per Second (Kibps) to Bytes Per Second (Bps). We can further simplify the formula to ease the calculation.
FORMULA
Bytes Per Second = Kibibits Per Second x 1024 / 8
STEP 1
Bytes Per Second = Kibibits Per Second x 128
Example : If we apply the above Formula and steps, conversion from 10 Kibps to Bps will be processed as below.
1. = 10 x 1024 / 8
2. = 10 x 128
3. = 1280
4. i.e. 10 Kibps is equal to 1,280 Bps.
(Result rounded off to 40 decimal positions.)
You can use above formula and steps to convert Kibibits Per Second to Bytes Per Second using any of the programming language such as Java, Python or Powershell.
#### Definition : Kibibit
A Kibibit (Kib or Kibit) is a binary unit of digital information that is equal to 1024 bits. It is defined by the International Electro technical Commission(IEC) and is used to measure the amount of digital data. The prefix "kibi" is derived from the binary number system, it is used to distinguish it from the decimal-based "kilobit" (Kb) and it is widely used in the field of computing as it more accurately represents the amount of data storage and data transfer in computer systems.
- Learn more..
#### Definition : Byte
A Byte is a unit of digital information that typically consists of 8 bits and can represent a wide range of values such as characters, binary data and it is widely used in the digital world to measure the data size and data transfer speed.
- Learn more..
### Excel Formula to convert from Kibps to Bps
Apply the formula as shown below to convert from Kibibits Per Second to Bytes Per Second.
ABC
1Kibibits Per Second (Kibps)Bytes Per Second (Bps)
21=A2 * 128
3
Download - Excel Template for Kibibits Per Second to Bytes Per Second Conversion
If you want to perform bulk conversion locally in your system, then download and make use of above Excel template.
### Python Code for Kibps to Bps Conversion
You can use below code to convert any value in Kibibits Per Second to Bytes Per Second in Python.
kibibitsPerSecond = int(input("Enter Kibibits Per Second: "))
bytesPerSecond = kibibitsPerSecond * 1024 / 8
print("{} Kibibits Per Second = {} Bytes Per Second".format(kibibitsPerSecond,bytesPerSecond))
The first line of code will prompt the user to enter the Kibibits Per Second as an input. The value of Bytes Per Second is calculated on the next line, and the code in third line will display the result. | 1,105 | 4,449 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2023-23 | latest | en | 0.823118 |
https://www.hitpages.com/doc/4510115092758528/10/ | 1,485,193,165,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560282935.68/warc/CC-MAIN-20170116095122-00491-ip-10-171-10-70.ec2.internal.warc.gz | 922,997,919 | 7,250 | X hits on this document
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BLUCK ET AL.
principal components analysis (PCA) because PCA originally was developed to produce a strong first, general (principal) factor. Since we expected three factors (not a general factor), we used EFA, which allows the variance to be better distributed across multiple factors. We also used EFA because PCA attempts to explain all of the vari- ance in a set of items (including unique and error variance), rather than only the reliable, common, shared variance among a set of items (Gorsuch, 1983).
Factors were extracted using a common factors model (principal axis) with a Promax (oblique) rotation in SPSS Version 11.0. We used an oblique rotation rather than the commonly used Varimax or- thogonal rotation procedure for two reasons. First, we had no reason to believe that the factors would be orthogonal. Previous discussions about the three theoretical functions of AM suggest that they may be intercorrelated (Bluck, 2003; Conway, 2003). In fact, it has been ar- gued that since most social science data involves correlated con- structs, oblique rotation procedures are preferable (Nesselroade, personal communication). Our inspection of the simple correlation matrix confirmed that, indeed, the items are intercorrelated. The sec- ond reason for using an oblique rotation procedure, particularly Promax, is that it produces a more simple structure than orthogonal or other oblique rotation methods, thereby increasing the ease of interpretation (Hendrickson & White, 1964).
In the initial EFA, Kaiser’s rule (Kaiser, 1960) of extracting factors with Eigen values > 1 suggested seven factors. Examination of the in- flection point on the scree plot suggested that six factors should be extracted (Cattell, 1966). These two solutions accounted for 63.12% and 59.25% of the variance, respectively. In addition, based on sam- ple size criteria delineated by Cliff and Hamburger (1967), a factor loading of .40 was used to identify meaningful factor loadings. Upon examining the factor pattern matrix of these solutions, not all factors were meaningful—two factors had only one or two item loadings greater than .40. Thus, a combination of Eigen value, scree inspec- tion, and loading criteria suggested either a four- or five-factor solution.
Next, because theory suggested the presence of three latent con- structs underlying responses to the TALE, we reran the EFA and forced three factors. The 3–factor solution, however, accounted for less than half the variance (44.15%). This solution was not ideal in that eight items did not load on any factor. Moreover, while the sec-
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# For Profit School Enrollments Drop 40% - Affecting MBA?
Author Message
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Kudos [?]: 24018 [1] , given: 4599
For Profit School Enrollments Drop 40% - Affecting MBA? [#permalink]
### Show Tags
23 Aug 2011, 22:48
1
KUDOS
Expert's post
Saw several articles covering for-profit colleges such as University of Phoenix, DeVry, and how enrollment numbers are severely down.
http://online.wsj.com/article/SB1000142 ... 01644.html
They cite several factors including the lack of ROI - I wonder if that is going to impact MBA programs since they are pretty much an ROI driven decision....
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Re: For Profit School Enrollments Drop 40% - Affecting MBA? [#permalink]
### Show Tags
24 Aug 2011, 10:07
bb wrote:
Saw several articles covering for-profit colleges such as University of Phoenix, DeVry, and how enrollment numbers are severely down.
http://online.wsj.com/article/SB1000142 ... 01644.html
They cite several factors including the lack of ROI - I wonder if that is going to impact MBA programs since they are pretty much an ROI driven decision....
I think it will impact some MBA programs. But i tend to belive that the higher the ranking the lower the impact will be.
Thanks for sharing news.
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Re: For Profit School Enrollments Drop 40% - Affecting MBA? [#permalink] 24 Aug 2011, 10:07
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Display posts from previous: Sort by | 829 | 3,079 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2017-13 | longest | en | 0.871811 |
http://everything.explained.today/Neighbourhood_(graph_theory)/ | 1,713,963,902,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819273.90/warc/CC-MAIN-20240424112049-20240424142049-00610.warc.gz | 11,154,245 | 5,956 | # Neighbourhood (graph theory) explained
In graph theory, an adjacent vertex of a vertex in a graph is a vertex that is connected to by an edge. The neighbourhood of a vertex in a graph is the subgraph of induced by all vertices adjacent to, i.e., the graph composed of the vertices adjacent to and all edges connecting vertices adjacent to .
The neighbourhood is often denoted or (when the graph is unambiguous) . The same neighbourhood notation may also be used to refer to sets of adjacent vertices rather than the corresponding induced subgraphs. The neighbourhood described above does not include itself, and is more specifically the open neighbourhood of ; it is also possible to define a neighbourhood in which itself is included, called the closed neighbourhood and denoted by . When stated without any qualification, a neighbourhood is assumed to be open.
Neighbourhoods may be used to represent graphs in computer algorithms, via the adjacency list and adjacency matrix representations. Neighbourhoods are also used in the clustering coefficient of a graph, which is a measure of the average density of its neighbourhoods. In addition, many important classes of graphs may be defined by properties of their neighbourhoods, or by symmetries that relate neighbourhoods to each other.
An isolated vertex has no adjacent vertices. The degree of a vertex is equal to the number of adjacent vertices. A special case is a loop that connects a vertex to itself; if such an edge exists, the vertex belongs to its own neighbourhood.
## Local properties in graphs
If all vertices in G have neighbourhoods that are isomorphic to the same graph H, G is said to be locally H, and if all vertices in G have neighbourhoods that belong to some graph family F, G is said to be locally F.[1] For instance, in the octahedron graph, shown in the figure, each vertex has a neighbourhood isomorphic to a cycle of four vertices, so the octahedron is locally C4.
For example:
• Any complete graph Kn is locally Kn-1. The only graphs that are locally complete are disjoint unions of complete graphs.
• A Turán graph T(rs,r) is locally T((r-1)s,r-1). More generally any Turán graph is locally Turán.
• Every planar graph is locally outerplanar. However, not every locally outerplanar graph is planar.
• A graph is triangle-free if and only if it is locally independent.
• Every k-chromatic graph is locally (k-1)-chromatic. Every locally k-chromatic graph has chromatic number
O(\sqrt{kn})
.
• If a graph family F is closed under the operation of taking induced subgraphs, then every graph in F is also locally F. For instance, every chordal graph is locally chordal; every perfect graph is locally perfect; every comparability graph is locally comparable; every (k)-(ultra)-homogeneous graph is locally (k)-(ultra)-homogeneous.
• A graph is locally cyclic if every neighbourhood is a cycle. For instance, the octahedron is the unique connected locally C4 graph, the icosahedron is the unique connected locally C5 graph, and the Paley graph of order 13 is locally C6. Locally cyclic graphs other than K4 are exactly the underlying graphs of Whitney triangulations, embeddings of graphs on surfaces in such a way that the faces of the embedding are the cliques of the graph.[2] Locally cyclic graphs can have as many as
n2-o(1)
edges.
## Neighbourhood of a set
For a set A of vertices, the neighbourhood of A is the union of the neighbourhoods of the vertices, and so it is the set of all vertices adjacent to at least one member of A.
A set A of vertices in a graph is said to be a module if every vertex in A has the same set of neighbours outside of A. Any graph has a uniquely recursive decomposition into modules, its modular decomposition, which can be constructed from the graph in linear time; modular decomposition algorithms have applications in other graph algorithms including the recognition of comparability graphs.
## References
see in particular pp. 89–90
• .
• .
• .
• .
• .
• .
• .
1. ,
2. ; | 879 | 4,009 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2024-18 | latest | en | 0.974845 |
https://school.quipper.com/en-PH/courses/math-grade-10-1/281-tree-diagrams-quiz.html | 1,529,705,657,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864822.44/warc/CC-MAIN-20180622220911-20180623000911-00277.warc.gz | 702,197,990 | 5,476 | # Probability using Tree Diagram
## Math - Grade 10 / Statistics and Probability
#### Sample Question
Two dice are thrown together.
Use a tree diagram to find the probability that one number is even and the other is odd.
• [% \frac {1}{2} %]
• [% \frac {1}{3} %]
• [% \frac {1}{4} %]
• 1
This is just one of our 121,230 study questions in Quipper School.
Quipper School Philippines Curriculum | 108 | 398 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2018-26 | latest | en | 0.84484 |
http://www.careerguide.co.in/2015/01/sbi-associate-bank-clerk-assistant_46.html | 1,545,208,110,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376831715.98/warc/CC-MAIN-20181219065932-20181219091932-00592.warc.gz | 344,198,849 | 16,262 | ### SBI Associate Bank Clerk/ Assistant Quantitative Aptitude/ Numerical Ability Model Test Paper 5 Solved Question Paper with Answer Key 2015
Here we have published 15 MCQs (Multiple Choice Questions) with answer key of Quantitative Aptitude/ Numerical Ability for SBI Associate Banks Clerk/ Assistant recruitment 2015 online examination. Hope you found it helpful and we wish you all the success for this exam.
### Quantitative Aptitude/ Numerical Ability Test Paper - 5 for SBI Associate Bank Clerk/ Assistant Exam 2015
Q.1. 4 men, 5 women and 3 children together can complete a piece of work in 16 days. In how many days can 10 women alone com-plete the piece of work if 10 men alone complete the work in 24 days ?
(A) 18
(B) 15
(C) 12
(D) Cannot be determined
(E) None of these
Q.2. What will be the difference bet-ween the simple interest and compound interest earned on a sum of Rs. 985 @ 14 p.c.p.a. at the end of two years ?
(A) Rs. 16•408
(B) Rs. 14•214
(C) Rs. 19•218
(D) Rs. 17•405
(E) None of these
Directions—(Q. 3 and 4) Study the information carefully to answer the questions that follow—
A basket contains 3 blue, 2 green and 5 red balls.
Q.3. If three balls are picked at random, what is the probability that at least one is red ?
(A) 1/2
(B) 7/12
(C) 11/12
(D) 1/5
(E) None of these
Q.4. If four balls are picked at random, what is the probability that two are green and two are blue ?
(A) 1/8
(B) 1/70
(C) 3/5
(D) 1/2
(E) None of these
Q.5. In how many different ways can the letters of the word ‘FLEECED’ be arranged ?
(A) 840
(B) 2520
(C) 1680
(D) 49
(E) None of these
Q.6. 2536+4851—?=3450+313
(A)3961
(B)4532
(C)3624
(D)4058
(E) None of these
Q.7. (2560 x 1.4) +(7400 x 0.6) =?
(A)7512
(B) 9746
(C)6523
(D) 8024
(E) None of these
Q.8. 36%of 850+? %of 592 = 750
(A)73
(B)89
(C)82
(D)75
(E) None of these
Q.9. 64%of 2650+40% 0f 320=?
(A)1824
(B) 1902
(C)1829
(D) 1964
(E) None of these
Q.10. 486+32×25—59=?
(A) 514
(B) 528
(C) 599
(D) 507
(E) None of these
Directions for (Q.11-13). There is a group of five teachers a,b,c,d and e
• b and c teaches maths and geography
• a and c teach maths and history
• b and d teach political science and geofraphy
• d and e tech political science and biology
• e teaches biology, history, and political science
Q.11. Who teaches political science, geography and biology
(A) e
(B) d
(C) c
(D) b
(E) None of these
Q.12. Who teaches mathematics, political science and geography:
(A) a
(B) b
(C) d
(D) e
(E) None of these
Q.13. Who teaches maths, geography and history
(A) c
(B) e
(C) a
(D) b
(E) None of these
Directions for (Q.14-15). Six members of a family are traveling these are a,b,c,d,e, and f. b is the son of c but c is not mothers of b. a and c are a married couple, e is the brother of c. d is the daughter of a. f is the brother of b.
Q.14. Who is the wife of e:
(A) a
(B) f
(C) b
(D) cant be determined
(E) None of these
Q.15. Which of following is pair of female
(A) ae
(B) bd
(C) df
(E) None of these
Q.1-
D
Q.2-
E
Q.3-
C
Q.4-
B
Q.5-
A
Q.6-
C
Q.7-
D
Q.8-
D
Q.9-
A
Q.10-
B
Q.11-
B
Q.12-
B
Q.13-
C
Q.14-
D
Q.15-
D
We have tried our best to present important questions with correct answer key. Any suggestion or update about any erroneous entry or answer is always appreciable. Please comment your score or any suggestion in comment section below.
## Want to Stay in Contact !
Get new posts and updates in your inbox. Subscribe through our Feedburner Email subscription. | 1,187 | 3,461 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2018-51 | longest | en | 0.843992 |
lsi-contest.com | 1,553,332,913,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202781.33/warc/CC-MAIN-20190323080959-20190323102959-00475.warc.gz | 121,674,257 | 4,678 | # 17th LSI Design Contests・in Okinawa Design Specification - 1
## 1. Noise Cancelling Algorithm
To remove noise, there are some methods are suggested such as Microphone Array Method which is used several microphones and method using adaptive filtering to presume unknown pathway. In this simulation, the Noise Cancelling System by using Specter Subtraction Method (SS Method) which removes the noise in frequency domain is targeted.
The difference from last year is that we introduce other efficient techniques described in Sec. 1-2.
## 1-1. Specter Subtraction Method (SS Method)
In this section, the Noise Cancelling Algorithm is explained. The algorithm removes the noise in frequency domain. A diagram of the Noise Cancelling Algorithm is shown in Figure 1.
Figure 1
First, the observed signal is assumed to be given by the sum of speech and noise. The observed signal, the speech and the noise at time n are defined as x(n), s(n) and d(n) respectively. Then, the observed signal x(n) is expressed by
when the speech and the noise are uncorrelated.
Next, to remove noise in frequency domain, the Discrete Fourier Transform (DFT) applies the observed signal x(n). After that, the observed signal specter X(l,k) from top to k in l-frame is expressed by
when S(l,k) and D(l,k) denote the speech specter and the noise specter respectively. The estimate value of speech is expressed by the observed signal specter X(l,k) as．
Here, G(l,k) is a specter gain, and the estimate value specter is result of the observed signal specter X(l,k) multiplied by suitable specter gain G(l,k). The ideal estimate value is expressed by
By substituting eq.(4) into eq.(3), the ideal specter gain G_opt (l,k) given as．
Then, the speech specter X(l,k) is extracted perfectly. However, the noise specter D(l,k) can’t be obtained by only the speech specter X(l,k)’s information. Therefore, for SS Method, the noise specter estimate value can be obtained by using the ‘L’ number of frames in non-speech interval. is defined as
The simulation of the SS method was performed by Scilab. The result is shown in Figure 2. We can download the program (here).
(a)Input Signal
(a)Output Signal
Figure 2
## 1-2. MAP method using Variable Speech Distribution
This is a new algorithm that we introduce this year. The file related to the algorithm was put in the DL_file_ver3 at here.
## 1-2-1. MAP Estimate Method
Object
Figure 3
・PDF to S and D
Suppose X=const.+D and no relation between D and S, then
Define ε in order to maximize S as
S can be rewritten as
Now we want to get following
From (16) and (17), we get
Therefore we get the spectrum gain
When we difine
・Decision-Directed Mehod
We often choose β=0.98 and γ(l) - 1 should be positive so that
・MAP Method
Most of speech signal follows Rayleigh distribution rather than Gaussian one
Figure 4
Suppose real and imaginary part of noise are uncorrelated each other with half of variance,
As same manner from (13) to (19), we get the spectrum gain for MAP estimate.
## 1-2-2. Speech spectrum distribution proposed by T.Lotter and P.Vary
PDF (Probability Density Function) of the speech spectrum that has been proposed by T.Lotter and P.Vary is one of the useful. According to them, the phase spectrum can be expressed approximately by an uniform distribution, the amplitude spectrum is also expressed approximately by the equation (7).
Γ(・) is the gamma function. ν and μ, is a parameter that determines the shape of the distribution. The figure 3 shows the PDF given by equation (26).
Figure 5
Figure 6
We get the spectrum gain for Lotter/Vary method
where
MAP estimated value of the phase spectrum and the phase spectrum of the observed signal.
## 1-2-3. Variable Speech Distribution
We show how to change voice PDF in Figure 5.
Figure 7
It is called Variable Speech PDF method that we adaptively change the shape of the speech spectrum distribution in the non-speech section and the speech section.
Equation (7) can be approximated to a Rayleigh distribution from exponential distribution by the parameter ν. We show the distribution curve which equation (7) gives on the case of each ν in Figure 6, and we fixed μ to 3.2.
Figure 8
When ν=0.0 Equation (7) match an exponential distribution, and when ν=2.0 (7) is approximated to a Rayleigh distribution. It means that we can approximate to the change of actual speechPDF by changing the value of ν.
The spectral gain based on the variable speech distribution is obtained by allowing variation in the spectral gain parameter of Lotter/Vary's.
The algorithm for determining ν(l,k) is given by
N is the number of FFT spectrum, and α is a parameter for adjusting the size of ν~(l,k).
The simulation of the MAP method was performed by Scilab. The result is shown in Figure 7. You can download the program here.
Figure 9
## References
[1] T. Lotter, P. Vary ; Speech enhancement by MAP spectral amplitude estimation using a super-Gaussian speech model, EURASIP Journal on Applied Signal Processing, 2005.
[2] A. Kawamura, M. Kurosaki ; 大容量化するマルチメディア・データを転送・保存・活用するために ディジタル音声&画像の圧縮/伸張/加工技術, in H. Ochi, CQ Publishing Co., 2013. | 1,274 | 5,157 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2019-13 | longest | en | 0.900614 |
http://www.numbersaplenty.com/33311918 | 1,585,482,707,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370494331.42/warc/CC-MAIN-20200329105248-20200329135248-00368.warc.gz | 292,989,077 | 3,422 | Search a number
33311918 = 2317332
BaseRepresentation
bin111111100010…
…0110010101110
32022200102100202
41333010302232
532011440133
63145553502
7553101203
oct177046256
968612322
1033311918
1117892812
12b1a5892
136b945c3
1445d1caa
152dd02e8
hex1fc4cae
33311918 has 12 divisors (see below), whose sum is σ = 51650208. Its totient is φ = 16096680.
The previous prime is 33311909. The next prime is 33311933. The reversal of 33311918 is 81911333.
It is a super-3 number, since 3×333119183 (a number of 24 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.
It is a congruent number.
It is an unprimeable number.
It is a polite number, since it can be written in 5 ways as a sum of consecutive naturals, for example, 45080 + ... + 45812.
It is an arithmetic number, because the mean of its divisors is an integer number (4304184).
It is a self-describing number.
Almost surely, 233311918 is an apocalyptic number.
33311918 is a deficient number, since it is larger than the sum of its proper divisors (18338290).
33311918 is an frugal number, since it uses more digits than its factorization.
33311918 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 1499 (or 766 counting only the distinct ones).
The product of its digits is 1944, while the sum is 29.
The square root of 33311918 is about 5771.6477716507. The cubic root of 33311918 is about 321.7608591803.
The spelling of 33311918 in words is "thirty-three million, three hundred eleven thousand, nine hundred eighteen". | 472 | 1,566 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2020-16 | latest | en | 0.871033 |
http://list.seqfan.eu/pipermail/seqfan/2005-November/006529.html | 1,716,659,020,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058830.51/warc/CC-MAIN-20240525161720-20240525191720-00109.warc.gz | 16,581,463 | 1,918 | # RE A problem sending a sequence (typo)
Tautócrona tautocrona at terra.es
Tue Nov 8 02:35:18 CET 2005
```----- Original Message -----
From: "Tautócrona" <tautocrona at terra.es>
Give an example:
a(2) = 5 because 5#-1 = 209 and 5#+1 = 211 are both prime at the same time, and 3 is the
only prime less than 5 having this property.
----------------------------
a(2) = 5 because 5#-1 = 29 and 5#+1 = 31 are both prime at the same time, and 3 is the
only prime less than 5 having this property.
Jose Brox
``` | 172 | 510 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2024-22 | latest | en | 0.881199 |
https://mathematica.stackexchange.com/questions/tagged/wavelet-analysis?tab=Active | 1,576,494,582,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541319511.97/warc/CC-MAIN-20191216093448-20191216121448-00448.warc.gz | 458,107,714 | 38,994 | # Questions tagged [wavelet-analysis]
Questions on the wavelet analysis functions of Mathematica.
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41 views
### Integral of a Wavelet function
I ran into problems integrating a function which contains both an If condition and a Wavelet which is an ...
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### <<WaveletsWavelets Not Found [closed]
please i'm a student and i need help in my project which is based on an already existing program. The problem that the program is ancient( Mathematica 5.0) and doesn't fit with the new versions of ...
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### WaveletListPlot: Broken in 11.2?
I'm trying to create a WaveletListPlot, following the reference code exactly: ...
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### Solve 1D Wave Equation
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### Numerical solution to approximate the singular integration using collocation method
I am working to solve "numerically" the following integral equation IE: u[x]=f[x]+Integrate[(1/(x-t)^(1/4))*u[t],{t,0,x}]+Integrate[(1/(x-t)^(1/4))*u[t],{t,0,1}] ...
140 views
### Integrating a function containing a wavelet
How do I evaluate the integral c? ...
157 views
### Computing FundamentalFrequency
How does Mathematica use AudioLocalMeasurements to compute the fundamental frequency? Is there any formula? I can't seem to find what method Mathematica makes use ...
29 views
### Multifractality - Wolfram Code [duplicate]
In order to descrive the multifractal behaviour of a set, Baumann provides a code based on the functions Dq and Tau, given below. ...
87 views
### Coefficients of continuous wavelet transform
I am fiddling with wavelet coefficients in Mathematica, and for simplicty only a real valued Wavlet. Lets say we have the data: ...
100 views
### Getting errors NIntegrate::nlim and NIntegrate::slwcon [closed]
Here is the modified question: First: Define phi[x_]:=Piecewise[{{1, 0 <= x < 1}}, 0] ...
85 views
### Revert image fusion using discrete wavelet transform
How can I extract a hidden image using discrete wavelet transform? I have used the function ImageFusion from this site to hide an image inside another one. ...
4k views
### Extracting information from the result of ContinuousWaveletTransform
I'm trying to analysis a signal and identify some time-frequency information of it. For example, around which time the specific frequency arrives. It appears that Mathematica has very powerful wavelet ...
148 views
### How to extract a single image from the output of the DiscreteWaveletTransform[]?
So, I use the following code to carry out the discrete wavelet transform on an image of Lena. I am interested in then doing some changes to a single image out of the obtained 4. How do I extract any ...
122 views
### How to detect the singular value with wavelet analysis
I have a series data data=Uncompress[ FromCharacterCode[Flatten[ImageData[Import["http://i.stack.imgur.com/ApipY.png"], "Byte"]]]]; We can show it ...
106 views
### wavelet fourier transform
I want to do wavelet Fourier transform of my data but I have few doubts in this. 1)first of all what is octive and voice command here when i am changing value according to that my plot is changing but ...
210 views
### Comparison of Discrete Cosine Transform and Discrete Wavelet Transform
I would like to compare the sorted coefficients of the Discrete Cosine Transform with that of the Discrete Wavelet Transform using the Haar wavelets. From theory, the discrete wavelet transform offers ...
188 views
### Wavelet filter artifact
I would like to denoise some data using wavelet filters. My current approach is to employ the stationary wavelet transform with Haar wavelets. However, this leads to undesirable artifacts at the top ...
2k views
### Wavelet vs instantaneous power spectrum
Following on from a previous question posted by @xslittlegrass and answered by @Sector and others, Extracting information from the result of ContinuousWaveletTransform I'd like to know if it is ...
158 views
### How to detect the singular value in 3-dimension with wavelet analysis
Actually this question is related with this post,but this is about point of 3-dimension ...
294 views
### How to detect discolorations in an image?
I'm evaluating the technical quality of an images and I'd like to scan it for defects from dirty sensors. Are there any filters to quickly detect a blob of discoloration? Here are a few examples:
1k views
### Detecting components in timeseries
I'm looking at this sequence of values: I'd like to detect the points where the center of the time-series shifts (around x=1000 and x=2000). Many of the transforms and smoothing methods I have tried ...
211 views
### How to calculate the amount of offset about two list
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The complex Morlet function is defined as: $$Ψ(t,f_c,f_b)= \frac{1}{\sqrt[]{ \pi f_{b} } }\exp(-t^2/f_b)\exp(\jmath 2πf_ct)$$ where $f_b$ and $f_c$ are two important parameters in modifying the ... | 2,615 | 11,330 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2019-51 | latest | en | 0.826976 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/165/6/v/ | 1,709,122,269,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474715.58/warc/CC-MAIN-20240228112121-20240228142121-00170.warc.gz | 881,667,289 | 54,407 | # Properties
Label 165.6.v Level $165$ Weight $6$ Character orbit 165.v Rep. character $\chi_{165}(38,\cdot)$ Character field $\Q(\zeta_{20})$ Dimension $928$ Sturm bound $144$
# Related objects
## Defining parameters
Level: $$N$$ $$=$$ $$165 = 3 \cdot 5 \cdot 11$$ Weight: $$k$$ $$=$$ $$6$$ Character orbit: $$[\chi]$$ $$=$$ 165.v (of order $$20$$ and degree $$8$$) Character conductor: $$\operatorname{cond}(\chi)$$ $$=$$ $$165$$ Character field: $$\Q(\zeta_{20})$$ Sturm bound: $$144$$
## Dimensions
The following table gives the dimensions of various subspaces of $$M_{6}(165, [\chi])$$.
Total New Old
Modular forms 992 992 0
Cusp forms 928 928 0
Eisenstein series 64 64 0
## Trace form
$$928 q - 8 q^{3} - 12 q^{6} + 140 q^{7} + O(q^{10})$$ $$928 q - 8 q^{3} - 12 q^{6} + 140 q^{7} - 2160 q^{10} - 1628 q^{12} + 1036 q^{13} + 1906 q^{15} + 53224 q^{16} + 122 q^{18} - 1992 q^{21} - 3400 q^{22} + 1032 q^{25} + 2866 q^{27} - 18892 q^{28} - 12506 q^{30} + 9656 q^{31} + 36278 q^{33} + 40196 q^{36} - 46264 q^{37} + 39812 q^{40} + 73094 q^{42} + 81920 q^{43} + 8612 q^{45} - 62904 q^{46} - 18746 q^{48} + 9868 q^{51} + 209772 q^{52} + 52888 q^{55} - 173730 q^{57} + 147052 q^{58} + 208390 q^{60} - 68304 q^{61} - 152414 q^{63} - 263420 q^{66} + 8448 q^{67} + 219340 q^{70} - 143068 q^{72} + 179360 q^{73} - 150766 q^{75} + 127312 q^{76} + 166820 q^{78} + 22708 q^{81} + 380572 q^{82} - 263532 q^{85} + 91164 q^{87} - 1922416 q^{88} + 1852576 q^{90} + 118336 q^{91} + 61372 q^{93} - 389656 q^{96} - 389256 q^{97} + O(q^{100})$$
## Decomposition of $$S_{6}^{\mathrm{new}}(165, [\chi])$$ into newform subspaces
The newforms in this space have not yet been added to the LMFDB. | 723 | 1,688 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-10 | latest | en | 0.310538 |
https://www.exceldemy.com/how-to-compare-two-cells-in-excel/ | 1,696,067,228,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510671.0/warc/CC-MAIN-20230930082033-20230930112033-00637.warc.gz | 824,537,320 | 64,002 | # How to Compare Two Cells in Excel (10 Easy Methods)
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There are many ways to compare cells in Excel. You can compare two cells and find matches, differences, and some other operations using Excel. In this article, we will discuss several easy and handy methods to compare cells.
Here, we show a sample method where we used the IF function to show whether there is any difference between the cells B5 and C5. The result is shown in the output column.
## 10 Easy Methods to Compare Two Cells in Excel
We are going to use the below dataset in order to demonstrate how you can compare two cell in Excel. to avoid any compatibility issue, you should opt for the Excel 365 edition.
### 1. Compare Two Cells Side by Side Using Equal to Sign
Using this method, you can find whether two cells contain similar data or not. Besides, this method compares data regardless of their type. For instance, we want to compare column Name 1 to column Name 2. Here are the following steps:
📌 Steps:
• Type the following formula in Cell D5 (to compare B5 & C5).
`=B5=C5`
• Drag down the Fill Handle (+) to copy the formula to the rest of the cells.
Read More: Compare Two Cells in Excel and Return TRUE or FALSE (5 Quick Ways)
### 2. Use IF Function to Compare Two Cells
Comparing using the IF function is very handy. Likewise, equal to sign, you can find matches/mismatches using this function. In our example, we will match column List 1 and column List 2.
📌 Steps:
• Insert IF function in Cell D5 and select the arguments.
`=IF(B6=C6,"Match","Not a Match")`
• Entered the above-mentioned formula you will get the following result. Finally, drag down the Fill Handle (+) of Cell D5 to copy the formula to the rest of the cells.
Read More: Return YES If 2 Cells Match in Excel (10 Methods)
### 3. Insert Excel EXACT Function to Compare Two Cells
Sometimes, cells contain text both in uppercase and in lowercase style. If you want to find if both the cells are containing uppercase or lowercase text then the EXACT function would be a great help.
📌 Steps:
• To compare Cell B5 and Cell C5, type EXACT function select necessary cells for arguments.
`=EXACT(B5,C5)`
• Upon entering the formula, you will get the following result. Later, drag down Fill Handle (+) of Cell D5 to copy the formula to the rest of the column.
Read More: Excel Compare Two Strings for Similarity (3 Easy Ways)
### 4. Combine IF and EXACT Functions to Compare Two Cells in Excel
You can compare two cells using the combination of IF and EXACT functions. Besides, the combination of these two functions is more effective. Because, the EXACT function checks the exactness of the data, and the IF function returns the condition of the data.
📌 Steps:
• To compare Cell B5 and Cell C5, enter the formula combining both functions:
`=IF(EXACT(B5,C5), "Match","")`
• After entering the formula, you will get the following result.
### 5. Highlight Matching Data to Compare Two Cells
Suppose, you have two different sets of data in excel and you want to analyze matched values in cells. Conditional formatting offers great help in solving such situations. Moreover, you can highlight matched cells very easily by using this method.
📌 Steps:
• Select the dataset.
• Go to Home > Conditional Formatting from Styles group.
• Click New Rule From Conditional Formatting.
• A new dialogue box will show up. Select the rule “Use a formula to determine which cells to format”.
• Type the following formula in the description box.
`=\$B5=\$C5`
• Click the Format button, go to the Fill tab and choose the color. Then, click OK.
Finally, If you follow the above steps correctly, all matched cells in the two columns will be highlighted. Conversely, differently named rows would not be highlighted.
Read More: Compare Two Cells Using Conditional Formatting in Excel (3 Methods)
### 6. Compare and Highlight Two Cells with Unique Data in Excel
Similar to the previous method, by using Conditional Formatting, you can compare two cells in various ways. For example, you can find unique values between two datasets.
📌 Steps:
• Select the dataset.
• Go to Home > Conditional Formatting from Styles group.
• Press the Duplicate Values option from Highlight Cell Rules.
• Subsequently, the Duplicate Values dialogue box will show up. Choose the Unique option from the drop-down.
• You can also choose the color of the highlight from the drop-down by using the Custom Format option.
• Then, enter OK.
• And, the final output is, all the unique values between cells are highlighted.
Read More: How to Compare Two Cells and Change Color in Excel (2 Ways)
### 7. Use LEFT & RIGHT Functions to Compare Two Cells Partially
Sometimes, you might need to match two cells partially. For example, you may need to compare only the first or last 3 characters of the cell. In those situations, LEFT or RIGHT functions can be used. The LEFT function returns the specified number of characters from the start of a text string. And similarly, the RIGHT function returns the characters from the right. In our example, we will match 3 Leftmost/Rightmost characters.
#### 7.1. Compare Using LEFT Function
📌 Steps:
• To match first 3 characters of Cell B5 and Cell C5, here is the formula using LEFT Function:
`=LEFT(B5,3)=LEFT(C5,3)`
• After entering the above formula correctly, the following is the output. Click Fill Handle (+) of Cell D5 to copy the formula to the rest of the column.
#### 7.2. Compare Using RIGHT Function
📌 Steps:
• To match the last 3 characters of cell H5 and cell I5, Insert the RIGHT Function and select or type arguments. Here is the Formula:
`=RIGHT(H5,3)=RIGHT(I5,3)`
• After you enter the above formula, the following is the output. Click the Fill Handle (+) of cell D5 to copy the formula to the rest of the column.
### 8. Using VLOOKUP and Find Matches in Excel
The VLOOKUP function is one of the easy ways to compare cells. It is commonly used to analyze Excel data. The VLOOKUP function looks for a value in the leftmost column in a table and then returns a value in the same row from the specified column. If you want to find any value to a column VLOOKUP function can be used.
📌 Steps:
• If we want to match the value of Cell C5 in Column Name 1, then the formula will be:
`=IFERROR(VLOOKUP(C5,\$B\$5:\$B\$11,1,0),"No Match")`
Breakdown of the Formula:
• VLOOKUP(C5,\$B\$5:\$B\$11,1,0)
Here, the VLOOKUP function looks for a value in the leftmost column of a table and then returns a value in the same row from a column you specify. So, the function will look for C5 in the range B5:B11 and return:
{John}
Conversely, when the function will find C6 in range B5:B11, it will return a #N/A error because C6 is not present in the prescribed range.
• IFERROR(VLOOKUP(C5,\$B\$5:\$B\$11,1,0), “No Match”)
The IFERROR function returns value_if_error if the expression is an error and the value of the expression itself otherwise. In our example, we have put “No Match” as an argument. As a result, when we will look for C6 in the above-mentioned range, the formula returns:
{No Match}
• After entering the formula, you will get the matched name in a 3rd column. Fill Handle (+) is used to copy the formula for the rest of the cells.
### 9. Using VLOOKUP and Find Differences
VLOOKUP can also be used to find differences between cells. the VLOOKUP function, in combination with IF & ISERROR function, finds a particular value in a range of data and returns the differences/similarity as output.
📌 Steps:
• If we want to find data in Cell C5 in Column Name 1, then the formula will be:
`=IF(ISERROR(VLOOKUP(C5,\$B\$5:\$B\$11,1,0)),"Not Available","Available")`
Breakdown of the Formula:
• VLOOKUP(C5,\$B\$5:\$B\$11,1,0)
The VLOOKUP function looks for a value in the leftmost column of a table and then returns a value in the same row from a column you specify. So, the function will return:
{John}
Unfortunately, this is not the ultimate result we want from this method. We want to know whether any value is present in a range or not. So, the next part of the formula is:
• ISERROR(VLOOKUP(C5,\$B\$5:\$B\$11,1,0))
Here, the ISERROR function checks whether a value is an error, and returns TRUE or FALSE. So, for D5, the function will find the value of C5 in the range B5:B11 and return:
{FALSE}
The reason is, C5 is present in the mentioned range. Likewise, for other cells when an error will be found, it will return “TRUE”.
• IF(ISERROR(VLOOKUP(C5,\$B\$5:\$B\$11,1,0)),”Not Available”,”Available”)
Now comes the final part. The IF function checks whether a condition is met, and returns one value if true, and another value if false. We put “Not Available” and “Available” as arguments. Finally, for D5, the function returns:
{Available}
📌 Step 2:
• After entering the formula you will find the differences in the Output Column. Fill Handle (+) is used to copy the formula for the rest of the cells.
### 10. Compare Two Cells with Greater Than or Less Than Criteria
Sometimes, you may need to compare two cells in excel to find which one is greater/lesser. For example, you can compare numbers, dates, etc. In such situations, we can use the IF function to do the comparison.
📌 Steps:
• In our dataset, if we want to compare between Cell B5 and Cell C5, We have used following formula:
`=IF(B5>C5,"Yes","No")`
• After entering the formula, here is the result. In our dataset, the date in Cell B5 is not greater than the date in Cell C5 so the output is No.
## Conclusion
There are many more ways to compare two Excel cells, but in this article, we tried to discuss easier methods. All these methods are easy to understand and take less time. If you have any questions regarding this article, please let us know.
## Related Articles
Hosne Ara
Hi, This is Hosne Ara. Currently, I do write for ExcelDemy. I have a long experience working with different industries and I have seen how vast the scope of Microsoft Excel is. So, eventually, I started to write articles on Excel and VBA. Basically, my articles are targeted to help people who are working in Excel. By profession, I am an Engineer. Materials and Metallurgical Engineering is my major. Besides, I am a certified Project Manager (PMP) too. I have worked with Power Plant and IT industry earlier. As a person, I am detail-oriented and love doing research. Establishing a greener world is one of my mottos.
1. In “10. Compare Two Cells with Greater Than or Less Than Criteria”, I would like to determine the actual numerical difference between the two cells and the percentage of that difference to one of the cells. Eg., if one cell is 2 and the other cell is 10 … the difference is 8, which is 80% less than 10 (10 would be the mean of my interested dataset). How to I go about writing the formula?
Thank you.
J
• Hi J,
Hope you are doing well.
You can get the actual numerical difference between two cells by using subtraction (-) in the formula. In case of, percentage value use the formula given below.
`=IF(AND(0<D5,D5<\$D\$10),(D5/\$D\$10)*100&"% Less Than 10",(-D5/\$D\$10)*100&"% Greater Than 10")`
If the condition is True then the formula will return the result as follows.
On the other hand, if the condition is False then the formula will return the result as follows.
Here, the AND function will return True if both the conditions are true else it will return False.
Thanks.
Regards,
Arin Islam
Advanced Excel Exercises with Solutions PDF | 2,789 | 11,510 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2023-40 | latest | en | 0.834428 |
https://askfilo.com/math-question-answers/find-the-number-of-tangents-to-the-curve-y2-2-x2-4-y80-which-pass-through-the | 1,726,240,429,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651523.40/warc/CC-MAIN-20240913133933-20240913163933-00292.warc.gz | 93,008,492 | 32,831 | Question
Easy
Solving time: 2 mins
# Find the number of tangents to the curve , which pass through the point .
A
0
B
1
C
2
D
3
## Text solutionVerified
Clearly, the point lies outside of the curve
. (as
Since the point lies outside of the given curve,
so the number of tangent will be 2 .
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Question Text Find the number of tangents to the curve , which pass through the point . Updated On May 11, 2023 Topic Application of Derivatives Subject Mathematics Class Class 12 Answer Type Text solution:1 Video solution: 2 Upvotes 271 Avg. Video Duration 5 min | 321 | 1,296 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-38 | latest | en | 0.835855 |
https://gorilla.bi/power-query/running-total-by-category/ | 1,713,060,653,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816863.40/warc/CC-MAIN-20240414002233-20240414032233-00109.warc.gz | 268,214,052 | 60,743 | Written By Rick de Groot
Rick is the founder of BI Gorilla. He believes learning is one of life's greatest pleasures and shares his knowledge to help you improve your skills.
Are you looking to create Running Totals by Category in Power Query? This article guides you through the process, step by step. We’ll use the List.Generate function and the Group By function to compute a grouped running total and make sure your calculations are both accurate and efficient.
If you’ve read my previous article on calculating running totals, you might be wondering how to extend that logic to categories. And how do you ensure good performance while doing so?
We’ll start by turning the running total logic into a reusable function. Then, we’ll use this function to calculate running totals for each category in your dataset.
## 1. What is a Grouped Running Total
A grouped running total is a running total that resets for each group in your dataset.
First, let’s clarify what a running total is. A running total is a way to keep track of a sum of numbers as they change over time. Essentially, it is a cumulative total of a series of numbers, where each new number is added to the previous total.
Now, imagine you’re tracking sales by region. A grouped running total would show you the sales for each region separately, resetting the total for each new region. This way, you can monitor how sales are evolving in each region, rather than just looking at the overall sales figure.
In technical terms, a grouped running total is like applying conditions to your running total. If you’re familiar with SQL, it’s similar to using the OVER and PARTITION BY clauses.
## 2. Running Total
Let’s start by revisiting how to create a basic running total. Then, we’ll adapt this into a reusable function.
### 2.1. Creating a Basic Running Total
Calculating a running total can be done in various ways. For the purpose of this article, we’ll focus on the List.Generate method for its speed and efficiency.
My other article starts with a query named “Sales”.
In the Running Total query, you’ll do the following:
• Use the List.Buffer function to load the values from the “Amount” column into memory.
• create a list that contains the running total values using List.Generate.
• Merge this list with your original table to include the Running Total values.
You can find all this in below code:
`````` let
Source = Sales,
BuffValues = List.Buffer( Source[Amount] ),
RunningTotal =
List.Generate (
() => [ RT = BuffValues{0}, RowIndex = 0 ],
each [RowIndex] < List.Count(BuffValues),
each [ RT = List.Sum( { [RT] , BuffValues{[RowIndex] + 1} } ),
RowIndex = [RowIndex] + 1 ],
each [RT] ),
#"Combined Table + RT" =
Table.FromColumns(
Table.ToColumns( Source )
& { Value.ReplaceType( RunningTotal, type {Int64.Type} ) } ,
Table.ColumnNames( Source ) & {"Running Total"} )
in
#"Combined Table + RT"``````
### 2.2. Converting the Logic into a Function
If you want to calculate running totals for different groups, it’s useful to convert this logic into a function. Here’s how:
1. Start by defining your variables at the beginning of the code.
2. Reference your table parameter, which we’ll call “MyTable,” throughout your code.
3. Use the Table.Column function to return a list of values.
You will end up with below code:
`````` ( RTColumnName as text, MyTable as table, ValueColumn as text) =>
let
Source = MyTable,
BuffValues = List.Buffer( Table.Column( MyTable, ValueColumn ) ),
RunningTotal =
List.Generate (
() => [ RT = BuffValues{0}, RowIndex = 0 ],
each [RowIndex] < List.Count(BuffValues),
each [ RT = List.Sum( { [RT] , BuffValues{[RowIndex] + 1} } ),
RowIndex = [RowIndex] + 1 ],
each [RT] ),
#"Combined Table + RT" =
Table.FromColumns(
Table.ToColumns( MyTable )
& { Value.ReplaceType( RunningTotal, type {Int64.Type} ) } ,
Table.ColumnNames( MyTable ) & { RTColumnName } )
in
#"Combined Table + RT"``````
Paste this code into the advanced editor and name your query: fxRunningTotal.
## 3. Running Total by Category
Understanding how to create a basic running total is a start, but what if you want to track running totals within specific categories? This chapter will guide you through the process.
### 3.1. Summarizing Data
The first step is to organize your data by the categories you’re interested in. In Power Query, you can do this using the Group By feature.
1. Navigate to the Home tab and click on Group By.
2. Choose the column that represents your category, such as “Product.”
3. Use the “All Rows” operation to summarize all the rows related to each category into a table object.
This will give you a unique row for each category, along with a table object containing all the related rows.
### 3.2. Understanding the Table.Group Function
Once your data is grouped by category, the next step is to apply the running total function to each of these groups. This is where the Table.Group function comes into play.
The Table.Group function is important for applying the running total to each category. Specifically, the third argument in this function, known as `aggregatedColumns`, is where the magic happens.
In our example, we use the “All Rows” operation, represented by an underscore (_), to collect all rows related to each category.
Here’s what the code looks like initially:
``````Table.Group(
#"Sorted Rows",
{"Product"},
{ { "Details",
each _,
type table [ Date = nullable date, Product = nullable text,
Amount = nullable number ]
} } )``````
### 3.2. Applying the Running Total Function
We’ve already created a function called `fxRunningTotal` that calculates a running total for a table. To apply this function to each category, wrap it around the underscore (_) in the `aggregatedColumns` argument.
Also, specify the name you want for the new Running Total column and the column that contains the values for calculating the running total.
We can also remove the data types specified, as the technique we use to combine the tables won’t need it.
Here’s how to modify the code:
``````Table.Group(
#"Sorted Rows",
{"Product"},
{ { "Details",
each fxRunningTotal( "Running Total", _, "Amount" ),
type table
} } )``````
After executing this code, you’ll find that each table object in the “Details” column now includes a Running Total column. To see this in action, click on the white space in a cell containing one of these table objects.
### 3.3. Finalizing and Combining Your Data
Typically, your next step would be to expand columns. However, instead of expanding the column, you can also use the Table.Combine function. This approach offers two main advantages:
1. It avoids hardcoding column names.
2. It automatically picks up the underlying data types of the columns.
The Table.Combine function requires a list of tables. Fortunately, we already have a column, “Details,” that contains table objects for each category. To create a list of these tables, you can use a column reference, also known as field selection.
If your last step was named “GroupedRows,” the following statement will create the list:
``GroupedRows[Details]``
This list contains tables that have been summarized and include the running total values.
To merge these tables into one, use the following code:
``Table.Combine( GroupedRows[Details] )``
After executing this code, you will find the columns expanded and including your Running Total by Category.
The method described here focuses on creating a running total by “Product,” but it’s flexible. You can easily adapt your Group By operation to categories like “Year and Month” to get a running total by month, or by “Week,” “Quarter,” or “Year.”
### 3.4. Consolidate Steps into Function
We build up the previous steps chronologically so it’s easier to understand how to compute a running total by category. And as you can see, it’s still quite a few steps.
If you are only interested in the results, you can also use the below function that combines the logic for both the running total and the summarizing and expanding of the data.
``````let func =
( RTColumnName as text, MyTable as table, ValueColumn as text, GroupByColumns as list ) =>
let
Source = MyTable,
//---------- RunningTotalFunction ----------
fxRunningTotal = ( RT_ColumnName as text, RT_Table as table, RT_ValueColumn as text) =>
let
Source = RT_Table,
BuffValues = List.Buffer( Table.Column( RT_Table, RT_ValueColumn ) ),
RunningTotal =
List.Generate ( () => [RT = BuffValues{0}, RowIndex = 0],
each [RowIndex] < List.Count(BuffValues),
each [ RT = List.Sum( { [RT] , BuffValues{[RowIndex] + 1} } ),
RowIndex = [RowIndex] + 1],
each [RT]
),
#"Combined Table + RT" =
Table.FromColumns(
Table.ToColumns( RT_Table )
& { Value.ReplaceType( RunningTotal, type {Int64.Type} ) },
Table.ColumnNames( RT_Table ) & { RT_ColumnName } )
in
#"Combined Table + RT",
//---------- END RunningTotalFunction ----------
// Applying function to GroupBy Operation
GroupedRows = Table.Group(
Source,
GroupByColumns,
{{"Details", each fxRunningTotal( RTColumnName, _, ValueColumn), type table }}
),
CombineData = Table.Combine( GroupedRows[Details] )
in
CombineData,
documentation = [
Documentation.Name = "fxGroupedRunningTotals",
Documentation.Description = " Adds a running total column to a table, based on a value column and one or more group by columns.",
Documentation.LongDescription = " This function adds a running total column to a table. For the running total you can specify the columns over which the running total should run. When you specify Year and Month as GroupByColumns, the running total will reset for each change in year and month in your dataset. It's important to sort your data in the desired order before you use this function. ",
Documentation.Category = " Running Total ",
Documentation.Source = " BI Gorilla â https://gorilla.bi ",
Documentation.Version = " 1.1 ",
Documentation.Author = " Rick de Groot ",
Documentation.Examples = {[Description = " ",
Code = " let
RTColumnName = ""Running Total"",
MyTable = Source,
ValueColumn = ""Amount"",
GroupByColumns = { ""Product"", ""Color""},
Result = fxGroupedRunningTotals( RTColumnName, MyTable, ValueColumn, GroupByColumns)
in
Result ",
Result = " Source Table that includes a running total by Product and Color
"]}]
in
Save this code as a separate query, and you can start using the function. Let’s say you have a table where the Amount column contains your values, and you want to create a running total column called Running Total and compute it by Color and Product. If you call this function fxGroupedRunningTotals, you can use it like this:
``````// Returns a table with a running total column by color and product.
fxGroupedRunningTotals( "Running Total", TableName, "Amount", { "Color", "Product" } )
// Returns a table with a running total column by product.
fxGroupedRunningTotals( "Running Total", TableName, "Amount", { "Product" } )
// Returns a table with a running total column
fxGroupedRunningTotals( "Running Total", TableName, "Amount", { } )``````
To see the code in action, you can download the accompanying file below.
## 5. Conclusion
And there you have it! You’ve learned how to calculate a running total by category in Power Query. We’ve covered everything from using the List.Generate function to using the Group By feature.
We began by creating a function to calculate a basic running total. From there, we delved into grouping your data and incorporating this function within the All Rows operation. Finally, we tackled how to combine the data back together. This comprehensive approach allows you to create running totals tailored to specific groups in your data.
## Tip
If you’re interested in further expanding your Power Query toolkit, you might want to explore how to create running total columns in bulk.
Enjoy Power Query!
Edit 05-10-2023: Improved function code to combine tables using Table.Combine.
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## Creating Julian Day Numbers with Power Query M
1. Hello Mr. Rick.
First of all, honestly thank you very much for your insightful article. I’ve spent a bunch of time to realize and internalize these code and the results were great. I’ve tried many ways and finally I can understand your codes.
But during my journey, I got a question. how should I change the code if I don’t want to call it as a function inside GroupBy? Indeed, I am curious to know if we can use RT Formula directly into GroupBy Function.
Best Regard
2. Hi Rick,
Thanks a lot for all your great posts and videos.
One suggestion for this Running Total by group function : it would be great if the function could also add an index column in the table at the “Group by” step.
I had to do further merges after the RT calculation and it changed the rows order, making the RT look erratic. An index column would make it super easy to reset the rows in the proper RT order.
3. Hi Rick,
Thank you for sharing such insightful blog. It is tremendously helpful to newbie like me.
Coming from Tableau background, I am struggling to find optimal performance solutions.
Can you recommend, how to modify the same use-case as provided in your example, running total for each product and date combination but with the condition that for each such combination, the running total column should be the cumulative of the previous 30 days (or maybe dynamic) value with respect to the current date value?
Thank you,
Ashish
4. Hi Rick, thank you so much for those explanations! they are really clear!
I am really struggling to decrement stock to allocate to orders on a SKU level.
I put the running totals, , but I cannot wrap my head around make the stock decremented by orders (considering orders are already sorted). please please please could you help me ????
thank you!
• What exactly is it that you want to achieve? It’s not completely clear to me.
5. Hi Rick,
Very helpful post and video. But I tried your code and it did not work on my data, until I figured out that the order of the columns is important. It only works when you put the Date as the first column of the table. It has probably to do with how the “group by” code works.
Anyway, thanks again!
• Hi Maarten,
I had the same problem and tried your solution and it works.
So, your comment was very useful.
Thanks!
6. Hi Rick,
Thank you for this incredibly helpful blog. Especially the step-by-step explanations in videos and text helped me to get started and to understand.
Have you perhaps already created an explanatory video/blog on how I can deal with data where the periods are filled unevenly? For example: Product A was sold in Jan,Feb,Mar, Product B only in Jan, Feb. If I now filter my table for March, I am missing the RT of Product B.
I probably need to create and link a full calendar for all periods in the background beforehand, but unfortunately I don’t know how exactly.
• Hey Daniel,
Thank you very much. Regarding your question, that is a common requirement. I don’t currently have a video on it, but if you need a running total by month, this is the general process:
1. Create a list of unique month-year combinations from the start to the desired end period you require. Isolating the column from a calendar, and removing duplicates works great.
2. Summarize your main transactions table by month-year combinations. If this table has only a date, add a month-year column, and then group your data. Make sure to SUM the underlying values in the group by operation.
3. Starting from the values from step 1, merge the values from step 2 with this table. You end up with a table that has all combinations for both products and year-months.
4. Perform the running total.
It’s a bit of a hassle in power query. Depending on your requirements and your host system, you could also do this in DAX.
Rick
7. Hi Rick,
Thank you for the clear video! I am using this for several workflows. I followed your instructions to create the function and when trying to invoke it on a specific one, I receive the following error:
An error occurred in the ââ query. Expression.Error: The field ‘Amount’ already exists in the record.
Details:
Name=Amount
Value=
If I rename the column to “Amount2” beforehand as a test, it says the same error for the new column name. I appreciate any assistance.
• Hi Marco – what parameters did you fill in for the function? The first argument provides the name of the running total column the function creates. Did you perhaps fill in ‘Amount’ there, although you already have a column named ‘Amount’?
• Hi Rick,
After taking another look in your detailed written steps, I noticed that the parameters were renamed.
( RTColumnName as text, MyTable as table, ValueColumn as text) =>
/( RT_Name as text, MyTable as table, RT_ColumnName as text) =>
I rewrote everything and this time I got the desired result. It seems I had a syntax error in the declarations that was not showing up and caused a loop. All good now. Great video!
8. Hi Rick,
Thank you for sharing this knowledge. Running totals have always been of great challenge and you’ve simplified it to a degree that is relatable and manageable to all. I had quite a complex operation to achieve with multiple “Grouped By” s and and a number of running totals (with if statements incorporated when running total reached a negative number there would be a reset to 0 and running total would continue).
• Chantal – That’s great to hear! And I agree it’s not an easy process at all. Hope they do something to improve this in the future.
9. you could improve the code by using Table.Combine,
namely you could replace:
``````#"Grouped Rows" =
Table.Group(
Source,
GroupByColumns,
{ {"Details", each fxRunningTotal ( RTColumnName, _, ValueColumn), type table } }
),
#"Removed Other Columns" =
Table.SelectColumns(#"Grouped Rows",{"Details"}),
#"Expanded Details" =
Table.ExpandTableColumn(#"Removed Other Columns", "Details",
Table.ColumnNames( Source ) & { RTColumnName },
Table.ColumnNames( Source ) & { RTColumnName } ),
RestoreDatatypes =
Value.ReplaceType(
#"Expanded Details",
Value.Type( // Creates dummy column to retrieve data type
Source,
RTColumnName,
each null,
Int64.Type
)
)
)
in ``````
by
``````group = Table.Group(
Source,
{"product"},
{{"Details", each fxRunningTotal( RTColumnName, _, ValueColumn), type table [Date=nullable date, Product=text, Amount=nullable number, running total=nullable number]}}
),
combine = Table.Combine( group [Details] )
in``````
• That’s a very good addition, also for respecting the data types. I have adjusted the article to include it. Thanks!
Rick
10. Hello Rick, thank you for this video it’s been enlightening.
Do you have any suggestions for summarizing two running totals at once? (Example: year to date budget and actuals) The buffer step doesn’t allow for multiple columns, should I repeat this process for the second column and merge the table results? Still new to PQ, but based on your video I’m guessing there’s a better solution.
11. Thanks for your blog articles, it helped me a lot !
I made an alternative version which identify before hands rows where running totals is reset. Before grouping I add a index to have grouping giving minimal index for each group. Then I transform it to a buffered list and use it to add or not previous running total.
Don’t know with a lot of data what is the most optimized. On my limited set it looked like it was the same. I also saw @StÊphane solution in comments for “Create Running Totals in Power Query M (Ultimate Guide)” which I tested and this one is definitely a lot slower.
Here my function :
``````(MyTable as table, ValueColumn as text, RTColumnName as text, optional GroupByColumns as list) =>
let
Source = MyTable,
#"Grouped Rows" = Table.Group(
GroupByColumns ?? {},
{{"Idx_Reset", each List.Min([Idx_Reset]), type number}}
),
BuffReset = List.Buffer(#"Grouped Rows"[Idx_Reset]),
BuffValues = List.Buffer(Table.Column(Source, ValueColumn)),
RunningTotal = List.Generate(
() => [RowIndex = 0, RT = BuffValues{0}],
each [RowIndex] < List.Count(BuffValues),
each [
RowIndex = [RowIndex] + 1,
RT = BuffValues{RowIndex} + (if not List.Contains(BuffReset, RowIndex) then [RT] else 0)
],
each [RT]
),
#"Combined Table + RT" =
Table.FromColumns(
Table.ToColumns(Source) & {Value.ReplaceType(RunningTotal, type {Currency.Type})},
Table.ColumnNames(Source) & {RTColumnName}
)
in
#"Combined Table + RT"``````
• After some test method to combine table take a lot of times imho.
I tested with a table of about 500 000 lines. Refresh was so long that I stopped it after 20 minutes.
I changed the way to retrieve running total inside the original table using Table.AddColumn and fetching for each row the corresponding value from running total list. Now it takes 8 minutes to run, still long but at least it finishes.
My new code (hoping it will format it correctly this time !?) :
`
``````(MyTable as table, ValueColumn as text, RTColumnName as text, optional GroupByColumns as list) =>
let
Source = MyTable,
#"Grouped Rows" = Table.Group(#"Index Added", GroupByColumns ?? {}, {{"Idx_Reset", each List.Min([Idx_Reset]), type number}}),
BuffReset = List.Buffer(#"Grouped Rows"[Idx_Reset]),
BuffValues = List.Buffer(Table.Column(Source, ValueColumn)),
RunningTotal = List.Generate(() => [RowIndex = 0, RT = BuffValues{0}],
each [RowIndex] < List.Count(BuffValues),
each [RowIndex = [RowIndex] + 1,
RT = BuffValues{RowIndex} + (if not List.Contains(BuffReset, RowIndex) then [RT] else 0)],
each [RT]),
BuffRunningTotal = List.Buffer(RunningTotal),
#"Combined Table + RT" = Table.RemoveColumns(#"RunningTotal Added",{"Idx_Reset"})
in
#"Combined Table + RT"``````
12. Hi Rick,
Thanks for great video and article!
I tried to filter on som dates after expanding the columns, step 3.3 with hope that it should run totals on selected dates only. It did not. What part should be dynamic to get wanted result?
• Can you clarify your desired result? I’m not entirely sure what it is you want to achieve.
13. Hi Rick,
Great series, extremely helpful. I was going through the running totals video and it has already helped in streamlining my process. Just wanted to check if you have any suggestion for my one issue. So I have Targets that are populated at weekly level however they are just duplicates as we have only Month level Targets. While creating the running total I need to get them on month level by just taking the average of week level Targets and add them on month level.
Would be great to know your thoughts on this one!
• Hi Venkatesh,
Sorry, it’s not entirely clear to me what you want for distributing the targets.
14. Hello Rick,
First and foremost I would like to thank you for your videos that are easy to follow and explain in detail what the individual lines do! This helped me tremendously!
I’ve been trying to calculate the amount of different currencies held in a specific portfolio at any given time. I could extract an Excel file of all the movements but it didn’t have a column showing the total currencies held. So I applied the different steps laid out by you and it works great! The only issue I have is that some of these currencies were exchanged between each other(i.e. 50âŦ exchanged to 50\$), these exchanges are referenced by one column listing the input currency, another the input currency amount, and the same for the output currency.
So I was thinking of simply adding a Running Total column for the Input Amount Currency and then using a Lookup function to
subtract from the Output Amount Currency. Is there a better way to do this in one go? Maybe a running total that adds and subtracts based on two different columns?
• Hey Arnaud,
This sounds like a great candidate to perform in DAX. Just have your values on each moment with the local currency. Then use some DAX to lookup the conversion.
You could also convert your values on a date level if you have the exchange rates for each. Perhaps merging a lookup table based on currency + date would be easiest.
15. I have spent years trying to figure out “Grouped Running Totals” in Power Query. I have visited this site many times, but it never clicked.
Today I watched the video (I usually just read the text) and everything made sense. This is so much easier than I thought, and I also learned a bit about functions.
I recommend that anyone who is struggling watch the video – so helpful. Thanks!
16. Hi Rick,
I’m on a third watch of this video and it just occurred to me…do we need the sort on the product? Doesn’t the Group By take care of that for us?
• Hey Deron,
Thanks for reaching out! I understand what you’re saying about the Product sort being unnecessary when you group your data by Product. It’s a great point and I appreciate you bringing it up.
When you group your data on Product, most likely the Product sort is unnecessary. After all, you will only apply the function on the rows belonging to that group.
17. Hello Rick,
I wonder want to know how to sum amount base on max date or buttom up,
I try to get change below sytanx in power query, but it didn’t work maybe there thing I miss. Please help me to direct the right way, Thanks!
``````List.Generate (
() => [ RT = BuffValues{Table.ColumnCount(MyTable)},
RowIndex = Table.ColumnCount(MyTable) ],
each [RowIndex] =0,
each [ RT = List.Sum( { [RT] , BuffValues{[RowIndex] -1} } ),
RowIndex = [RowIndex] â 1 ],
each [RT] )``````
• I think I made it work, but unfortunately i think i not really know about list.generate this fun.
``````RunningTotal =
List.Reverse(
List.Generate (
() => [ RT = BuffValues{List.Count(BuffValues)-1},
RowIndex = List.Count(BuffValues)-1],
each [RowIndex]>=0,
each [ RT = List.Sum( { [RT] , BuffValues{[RowIndex] -1} } ),
RowIndex = [RowIndex] â 1 ],
each [RT] ))``````
18. Hi Rick,
Your detail documents and videos are the best of the best.
By the way, I have bank statements which have many transaction dates, and also different columns for Debit Amount, and Credit Amount.
I need to find a way to produce Running Total of Debit Amount Column, and another Running Total for Credit Amount Column.
Then I will be able to produce daily closing balance.
• Sovan – you could create a running total (using list.generate) that computes a debit column, and one that creates a credit column.
Those would be 2 separate statements.
And for the credit one you would add an if condition to only deduct a value when it’s < 0. For the debet one the opposite. Finally you would have to add 2 columns to the original table, not one. It can be done with similar logic đ Cheers, Rick
19. Hi Rick,
Your blogs have been extremely helpful! Especially your two running total ones. I was wondering, is there was a way to restart a running total after it hits a certain value? I am tracking values that are going to be adding together and should be negative, but if the running total goes above 0, I want that value to be shown in the row it happens and the next row to start running total over. Is there a way to do this?
• Blake – You should be able to add the running total logic you describe in List.Generate’s 3rd argument. It would require checking whether the RT value is >= 0.
Something like:
Then depending on what you want to return (do you want to return everything that is > 0 in the new running total, or just discard it).
``````RT = if List.Sum( { [RT] , BuffValues{[RowIndex] + 1} } ) > 0
then x
else List.Sum( { [RT] , BuffValues{[RowIndex] + 1} } )``````
It would all be the conditional logic. It sounds quite a challenge. I don’t have a ready made example for you cooked up, but hope you can play around with it.
• Hi Rick & Blake,
Thank you very much Rick, extremely helpful. I have a similar request to Blake except I want the Running Total to reset to 0 when the Amount is 0. My Use Case is I am calculating the number of consecutive working day shifts for each employee, but once the employee does not work the shift that day then the Running Total should reset to 0. Would you please show me how to do this?
• Hi Warren. I would suggest an alternative approach.
If you sort your data in the correct way (probably datetime and employee), you can group your data by EmployeeID using GroupKind.Local. Also see:
https://powerquery.how/groupkind-local/
Then you can add an index column to each of those groups shown here:
In that way you don’t need a complex List.Generate statement.
However, I understand there’s more people having the need to reset a list.generate statement based on a statement. I’ll consider writing an article on it.
Cheers, Rick
• Hi Rick,
Thank you for your feedback. I applied the changes you suggested and hasn’t quite solved the challenge I face. Please assume in my data I have 30 days of shifts for each employee. With your changes I now have an index starting from 1 – 30 for each employee which is great. If employee#1 worked days 1 – 3, the “Consecutive Day Shift Count” is correctly showing the running total count of 3 on day 3. However, on day 4 the employee did not work (“Shift Count” = 0) so instead of resetting the count to 0 on day 4; the “Consecutive Day Shift Count” is still showing the previous running count of 3. It will stay at 3 until the next working shift when it moves to 4. My aim is to reset the “Consecutive Day Shift Count” to 0 on Day 4. Then on day 5 if the employee works again the “Consecutive Day Shift Count” should be 1. I hope this helps explain my challenge. I thank you again for you help.
• Can it be that you haven’t applied GroupKind.Local? The issue you describe happens when applying GroupKind.Global.
If you add a flag that says: Hasworked (true or false) and include that in the grouping, then the consecutive series is captured by Groupkind.Local. This means the running index resets for each series.
20. Hi Rick,
Could you please let me know if this would work for a running total across 2 or more categories? For example, to create a running total by product and month, would I need to create individual product tables and then compile after creating the running total calc?
Thanks
• Yes, because you can Group By more than one column. In my case I sorted and then grouped by six different columns. It all worked a treat!
• Hi Joe,
Chris is right. You can perform the Group By operation on any number of columns you want (in your case on Product and Month).
Then call the function on the table objects that have been summarized on that level, and expand.
You should then have your desired results đ
21. Hi Rick,
Thank you for you great job!
One minor thing: is there some logic behind your parameters order?
Just in all powerquery functions that I know (Table.AddColumn, for instance), the first argument is TableName (in 99% cases, it’s previous step name), and the ColumnName is the second, not vice versa. | 7,057 | 30,839 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2024-18 | latest | en | 0.906691 |
https://www.reading.ac.uk/modules/document.aspx?modP=ICM292&modYR=2021 | 1,604,034,554,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107907213.64/warc/CC-MAIN-20201030033658-20201030063658-00062.warc.gz | 856,532,933 | 5,379 | ## ICM292-Derivatives Modelling
Module Provider: ICMA Centre
Number of credits: 20 [10 ECTS credits]
Level:7
Terms in which taught: Spring term module
Pre-requisites:
Non-modular pre-requisites:
Co-requisites: ICM127 Stochastic Calculus and Probability
Modules excluded:
Current from: 2020/1
Module Convenor: Dr Emese Lazar
Email: e.lazar@icmacentre.ac.uk
Type of module:
Summary module description:
This module introduces the main approaches used for derivatives pricing, based on the concepts covered in the Stochastic Calculus and Probability module. It discusses discrete time as well as continuous time valuations, including the Black-Scholes model and the martingale approach. These ideas are developed further in the Advanced Derivatives Modelling module, whilst this module provides a link with the Numerical Methods for Financial Engineering module as well.
Aims:
To convey the basic concepts and analytical methodology for the valuation of derivatives in the standard Black-Scholes framework.
Assessable learning outcomes:
By the end of the module, it is expected that the student will be able to:
• derive the price, in discrete and continuous frameworks, using different methods, for a variety of equity based simple and exotic derivatives
• digest the literature on equity based derivatives at an intermediary level, compare different methodologies and evaluate results
The module creates awareness of the mathematical foundation for working in the area of financial derivatives’ pricing. This will also create motivation and background for further study in other areas as well (eg. the pricing of interest rate and credit derivatives). The students will get an introduction into the models and pricing of interest rate and credit derivatives.
Outline content:
1.Introduction, use of derivatives, the greeks
2.Discrete time valuation
3.Continuous time valuation
4.Black-Scholes model, properties and extensions
5.Martingale approach
6.Complete and incomplete markets
7.Claims on currencies and multiple assets; foreign equity markets
8.Selected equity, interest rate and credit derivatives
Brief description of teaching and learning methods:
Teaching is based on tailor made lecture notes.
Compulsory homework assignments are set weekly.
Lectures are supported by discussions of the homework assignments in interactive seminars.
Contact hours:
Autumn Spring Summer Lectures 20 Seminars 10 Guided independent study: Wider reading (independent) 60 Exam revision/preparation 40 Advance preparation for classes 10 Preparation for presentations 10 Revision and preparation 20 Essay preparation 25 Reflection 5 Total hours by term 0 200 0 Total hours for module 200
Summative Assessment Methods:
Method Percentage Written exam 60 Written assignment including essay 20 Class test administered by School 20
Summative assessment- Examinations:
One written final exam (closed book) of length 2 hours.
Summative assessment- Coursework and in-class tests:
5 written assignments (take home, open book) with submission dates in weeks 4, 6, 7, 8 and 9.
One class test (open book) of length 1 hour 30 minutes.
Formative assessment methods:
Penalties for late submission:
Penalties for late submission on this module are in accordance with the University policy. Please refer to page 5 of the Postgraduate Guide to Assessment for further information: http://www.reading.ac.uk/internal/exams/student/exa-guidePG.aspx
Assessment requirements for a pass:
50% weighted average mark
Reassessment arrangements:
By written examination only, to be taken in August/September, as part of the overall examination arrangements for the MSc programme | 756 | 3,660 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2020-45 | latest | en | 0.86268 |
https://algebra-calculator.com/algebra-calculator-program/radicals/tutorial-on-permutation-and.html | 1,726,672,444,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651899.75/warc/CC-MAIN-20240918133146-20240918163146-00098.warc.gz | 73,431,415 | 10,953 | Algebra Tutorials!
Home Solving Quadratic Equations by Completing the Square Graphing Logarithmic Functions Division Property of Exponents Adding and Subtracting Rational Expressions With Like Denominators Rationalizing the Denominator Multiplying Special Polynomials Functions Solving Linear Systems of Equations by Elimination Solving Systems of Equation by Substitution and Elimination Polynomial Equations Solving Linear Systems of Equations by Graphing Quadratic Functions Solving Proportions Parallel and Perpendicular Lines Simplifying Square Roots Simplifying Fractions Adding and Subtracting Fractions Adding and Subtracting Fractions Solving Linear Equations Inequalities in one Variable Recognizing Polynomial Equations from their Graphs Scientific Notation Factoring a Sum or Difference of Two Cubes Solving Nonlinear Equations by Substitution Solving Systems of Linear Inequalities Arithmetics with Decimals Finding the Equation of an Inverse Function Plotting Points in the Coordinate Plane The Product of the Roots of a Quadratic Powers Solving Quadratic Equations by Completing the Square
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Fard
Registered: 10.09.2004
From: Texas
Posted: Tuesday 26th of Dec 12:07 I'm just wondering if someone can give me a few pointers here so that I can understand the concepts behind tutorial on permutation and combination e book. I find solving problems really difficult. I work in the evening and thus have no time left to take extra tutoring . Can you guys suggest any online resource that can assist me with this subject?
AllejHat
Registered: 16.07.2003
From: Odense, Denmark
Posted: Wednesday 27th of Dec 11:12 I think I know what you are searching for. Check out Algebrator. This is an amazing tool that helps you get your homework done faster and right. It can assist you with problems in tutorial on permutation and combination e book, lcf and more.
Ashe
Registered: 08.07.2001
From:
Posted: Wednesday 27th of Dec 17:50 Algebrator is the perfect algebra tool to help you with projects. It covers everything you need to know about roots in an easy and comprehensive way . Math had never been easy for me to grasp but this software made it easy to comprehend . The logical and step-by–step approach to problem solving is really an advantage and soon you will find that you love solving problems.
medaxonan
Registered: 24.08.2003
From:
Posted: Thursday 28th of Dec 13:09 I understand. My concepts are quite clear, but this specific set seems to be very difficult . A little help would do me a lot. Please give me the URL to it.
Svizes
Registered: 10.03.2003
From: Slovenia
Posted: Thursday 28th of Dec 17:34 I am a regular user of Algebrator. It not only helps me finish my homework faster, the detailed explanations offered makes understanding the concepts easier. I suggest using it to help improve problem solving skills.
malhus_pitruh
Registered: 23.04.2003
From: Girona, Catalunya (Spain)
Posted: Friday 29th of Dec 11:19 It is available at https://algebra-calculator.com/recognizing-polynomial-equations-from-their-graphs.html and really is the easiest program to get up and running. You can start learning math within minutes of downloading the software. | 993 | 4,215 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2024-38 | latest | en | 0.832865 |
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## #1 2013-02-25 21:13:42
panicoschr
Member
Offline
### Standard Deviation
Hello. Can someone help me on this ?
Is it possible to calculate Standard Deviation for points distrubuted in 3-axis?
X Y Z.
Thanks
## #2 2013-02-25 21:20:24
bobbym
Offline
### Re: Standard Deviation
Hi;
Welcome to the forum. You could get the standard deviation of the x's, then the y's and then the z's.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #3 2013-02-25 22:35:12
bobbym
Offline
### Re: Standard Deviation
Standard deviation like a mean works on a list of elements not on 3 at a time.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #4 2013-03-22 01:35:29
panicoschr
Member
Offline
### Re: Standard Deviation
Hello again
i have another issue, if you could assist.
I have an array of integers from 0 - 250 positions, each one containing an integer. Is it possible to get a single value for this array, so that I can tell the difference between this array and other future arrays? Perhaps an integral?
## #5 2013-03-22 01:39:17
bobbym
Offline
### Re: Standard Deviation
There are several ideas but I would need to know a little bit more.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #6 2013-03-22 02:10:16
bobbym
Offline
### Re: Standard Deviation
Hi;
You are saying that you are using an array to keep track of 251 colors. You are assigning each box to a color and incrementing it by one each time you find one?
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #7 2013-03-22 02:19:42
bobbym
Offline
### Re: Standard Deviation
Okay, what are you trying to do with that? Do you have some other array to compare it to?
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #8 2013-03-22 02:26:52
panicoschr
Member
Offline
### Re: Standard Deviation
it would be easy to compare it with another array, slot by slot. But i want to get a value out of each array. Thinking of the array as a graph-> can i get a value out of it? area perhaps?distance of the area from 0? perhaps 2 or 3 small areas on a graph separated by 0 color values? sorry i can not think anything to compare 2 graphs that may look completely different.
## #9 2013-03-22 02:35:11
bobbym
Offline
### Re: Standard Deviation
You want to turn the array into a single number? This could very well take much more time than comparing them element by element.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #10 2013-03-22 02:37:17
panicoschr
Member
Offline
### Re: Standard Deviation
can i do this, turn it to a number?
## #11 2013-03-22 02:44:18
bobbym
Offline
### Re: Standard Deviation
You are using some programming language?
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #12 2013-03-22 02:46:22
panicoschr
Member
Offline
### Re: Standard Deviation
yes I am using C++ in eclipse.
## #13 2013-03-22 02:49:30
bobbym
Offline
### Re: Standard Deviation
Does the new C++ have multiprecision arithmetic yet?
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #14 2013-03-22 02:54:03
panicoschr
Member
Offline
### Re: Standard Deviation
I don't know what is this.
But i know it has fields like float and double.
Last edited by panicoschr (2013-03-22 02:54:25)
## #15 2013-03-22 02:56:06
bobbym
Offline
### Re: Standard Deviation
It still only has 16 digits of precision is what I mean. Basically that means only 16 digits in any number.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #16 2013-03-22 03:14:59
bobbym
Offline
### Re: Standard Deviation
No, not exactly. Remember order counts in your array so you just can't add up the numbers or multiply them. The simplest thing of course is the decimal system. How high are the numbers in the each array?
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #17 2013-03-22 05:42:31
panicoschr
Member
Offline
### Re: Standard Deviation
sorry i had to leave for some time. I made a small testing , saw max no = 160. Of course most of the slots are value 0. Lets say only 30 slots have a value {3,4, 10 , 20 , 34 etc)
## #18 2013-03-22 09:40:17
bobbym
Offline
### Re: Standard Deviation
Hmmm, that is going to be a problem.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof. | 1,697 | 6,245 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2014-15 | longest | en | 0.904736 |
https://yuihuang.com/atcoder-abc-169e/ | 1,721,532,073,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517550.76/warc/CC-MAIN-20240721030106-20240721060106-00468.warc.gz | 911,702,027 | 15,790 | # 【題解】AtCoder ABC 169E – Count Median
```#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int n, a, b;
vector <int> l, r;
int main() {
cin >> n;
for (int i = 0; i < n; i++){
cin >> a >> b;
l.push_back(a);
r.push_back(b);
}
sort(l.begin(), l.end());
sort(r.begin(), r.end());
if (n % 2 == 1){
cout << r[n/2]-l[n/2]+1 << "\n";
}
else{
cout << (r[n/2]+r[n/2-1])-(l[n/2]+l[n/2-1])+1 << "\n";
}
}
``` | 168 | 429 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-30 | latest | en | 0.340939 |
cagleservice.com | 1,721,067,709,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514713.2/warc/CC-MAIN-20240715163240-20240715193240-00383.warc.gz | 134,719,207 | 40,617 | What is AFUE for Furnaces? - Cagle Service Heating & Air | Jackson TN
# What is AFUE for Furnaces?
When it comes to understanding the numbers and traits of a furnace, there is probably nothing more important than AFUE or the AFUE rating. In short, this term determines the efficiency of the furnace. The AFUE rating is similar to the SEER or EER rating for an air conditioner. “AFUE” stands for Annual Fuel Utilization Efficiency. It corresponds to how efficiently the furnace utilizes and uses it’s fuel – whether it is propane, natural gas, electric or oil. The most efficient furnaces use the smallest ratio of usable fuel to wasted fuel that is needed to heat up the targeted space.
But how exactly is it calculated? And what is a good AFUE?
Let’s get into those questions below!
## How is AFUE Calculated?
Annual fuel utilization efficiency is calculated by determining the percentage of fuel that is actually used in producing the heat that is applied to the targeted area in your home or business. In other words, it states how much of the fuel is actually applied towards heating the indoor environment. It states how much of the fuel is being utilized.
Wait, what? Not all of the fuel is utilized?
What many people don’t know is that the fuel that a furnace uses to produce heat usually isn’t entirely applied to heating the inside of your home or business. In most cases, a portion of it is wasted. For example, a standard furnace uses about 80% of the heat produced to heat the indoors. The remaining 20% of that fuel that is used is lost during the process. Most of it exits through the exhaust. Higher efficiency furnaces can have an efficiency of up to 98.5%. This would mean that only around 1.5% of fuel is wasted.
There is one type of furnace that is 100% efficient and has a AFUE rating of 100%, and that is an electric furnace. We will get more into electric furnaces a little later.
Now that we’ve gone over what annual fuel utilization efficiency is and how it’s calculated, let’s get into what a good, preferred rating would be.
## What is a Good AFUE Rating?
As it is with most subjective topics, determining a good annual fuel utilization efficiency rating is going to come down to opinion. And we are happy to share ours!
We believe the standard mid-range rating of 80% – 85% is a “good” rating. Is it the best? No, but it is good and it should be sufficient for most applications. The best ratings are going to be high efficiency furnaces which range from 90% – 98.5%. These furnaces will cost more upfront, but they will typically end up making up the difference through energy savings in the long run. Then, of course, we have electric furnaces which are the most efficient of all. Their rating of 100% is tops, but there are some other things to consider, especially if savings is your goal.
## Some Factors to Consider
The AFUE rating is a great tool to use when shopping for a new furnace. But, if it is savings you are after, don’t consider this rating only. It’s important to consider the type of furnace you are wanting or needing – to be specific, the type of fuel it uses. For example, if you want an electric furnace because of it’s spectacular annual fuel utilization efficiency rating of 100%, don’t let that be your only factor to consider. In most places, electricity costs more than gas. So, even though an electric furnace has the best efficiency rating, it still might cost you more to run the unit over the course of a year because it’s fuel source costs more than the alternatives.
In short, annual fuel utilization efficiency ratings are best used when comparing the same types of furnaces. For example, comparing propane furnaces with propane furnaces. When it comes to savings, comparing different types of furnace’s AFUE ratings can be misleading. As stated above, the fuel source is also going to affect the cost of operation. Both factors need to be considered.
You will also want to consider the current fuel type of the application. Is your specific application all electric? Or does it already have natural gas lines that are accessible? Sometimes, switching to a cheaper fuel source isn’t worth the initial upfront costs of making the switch.
## I Hope This Has Helped!
Remember, AFUE simply refers to the efficiency rate that the furnace uses it’s fuel source. The higher the rating, the more fuel is actually utilized and less is wasted. When comparing annual fuel utilization efficiency ratings, make sure you are comparing furnaces with the same fuel source. And lastly, also consider the cost of the fuel source and the initial upfront costs of switching to a new fuel source.
Related: 4 Tips for Buying a New Furnace | 1,003 | 4,698 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-30 | latest | en | 0.958486 |
https://www.teacherspayteachers.com/Product/Multiplication-Galore-2896830 | 1,516,428,128,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889325.32/warc/CC-MAIN-20180120043530-20180120063530-00225.warc.gz | 982,112,719 | 15,107 | Total:
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# Multiplication Galore
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This unit contains hands on activities to teach multiplication! They are engaging and low prep!
*Spin A Fact (Multiplication Game)
*Let's Multiply- Students use a deck of cards to build multiplication equations and find the products.
*Multiplication Four In a Row (3 Games)
*Multiplication Match Up- Students Match the equation to the correct picture.
*Multiplication Word Problems- Students solve the word problems on the task cards.
*Multiplication Choice Boards- ( 4 Boards) Mixture of multiplication and comparison problems.
*Multiplication Word Problem Work Boards- Students can practice solving word problems while showing that they've demonstrated how to use strategies to solve.
*Model Word Problems- Students have a step by step guided to help them solve tricky word problems.
*Multiplication Tic Tac Toe- One digit, one digit by two digit, and two digit by two digit
*Estimate the Product Practice Pages
*Box Method Printables=
*What's the equation?- Students use a bubble map to identify the equation for the given multiples
*Multiplication Drill Practice
*Missing Multiples- Students practice finding the multiples! Great for homework practice!
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\$7.50 | 338 | 1,446 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2018-05 | latest | en | 0.81065 |
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#### Online Quiz (WorksheetABCD)
Questions Per Quiz = 2 4 6 8 10
### Grade 1 - Mathematics4.4 Money Story Problems
Money Facts: 1 penny = 1 cent 1 nickel = 5 cents 1 dime = 10 cents 1 quarter = 25 cents 1 dollar = 100 cents Directions: Answer the following questions. Also write at least ten story problems of your own.
Q 1: How much is four pennies, one quarter, and one dime?40 cents39 cents10 cents Q 2: If I have four nickels and buy gum that costs 15 cents, how much money did I spend?10 cents15 cents1 cent Q 3: Sam has 30 cents and Dante has 26 cents. How much more money does Sam have than Dante?1 pennyfour penniesfive pennies Q 4: Sam buys a coke for 55 cents. Which set of coins does she need to buy the coke?5 dimes, 5 pennies1 nickel, 2 dimes, 4 pennies2 quarters, 1 penny Q 5: Jan has 25 cents and buys a toy that costs 10 cents. How much money does Jan have left?15 cents25 cents10 cents Q 6: Sam has 55 cents and buys an apple for 30 cents. How much money does she have now?two dimesfive pennies1 quarter Q 7: Which set equals 37 cents?5 nickels, 2 pennies7 nickels, 2 pennies2 pennies, 1 dime, 5 quarters Q 8: Tom has 30 cents and Joe has 20 cents. How much more does Tom have than Joe?10 cents20 cents`40 cents Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only! | 419 | 1,453 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2018-43 | latest | en | 0.94731 |
http://kintali.wordpress.com/category/graph-theory/ | 1,419,041,430,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802769321.94/warc/CC-MAIN-20141217075249-00114-ip-10-231-17-201.ec2.internal.warc.gz | 156,274,194 | 20,025 | # Directed Minors III. Directed Linked Decompositions
This is a short post about the following paper in my directed minor series :
• Shiva Kintali. “Directed Minors III. Directed Linked Decompositions“. Preprint available on my publications page.
Thomas [Tho’90] proved that every undirected graph admits a linked tree decomposition of width equal to its treewidth. This theorem is a key technical tool for proving that every set of bounded treewidth graphs is well-quasi-ordered. An analogous theorem for branch-width was proved by Geelen, Gerards and Whittle [GGW’02]. They used this result to prove that all matroids representable over a fixed finite field and with bounded branch-width are well-quasi-ordered under minors. Kim and Seymour [KS’12] proved that every semi-complete digraph admits a linked directed path decomposition of width equal to its directed pathwidth. They used this result to show that all semi-complete digraphs are well-quasi-ordered under “strong” minors.
In this paper, we generalize Thomas’s theorem to all digraphs.
Theorem : Every digraph G admits a linked directed path decomposition and a linked DAG decomposition of width equal to its directed pathwidth and DAG-width respectively.
The above theorem is crucial to prove well-quasi-ordering of some interesting classes of digraphs. I will release Directed Minors IV soon. Stay tuned !!
# Directed Width Parameters and Circumference of Digraphs
This is a short post about the following paper :
• Shiva Kintali. “Directed Width Parameters and Circumference of Digraphs“. Preprint available on my publications page.
We prove that the directed treewidth, DAG-width and Kelly-width of a digraph are bounded above by its circumference plus one. This generalizes a theorem of Birmele stating that the treewidth of an undirected graph is at most its circumference.
Theorem : Let G be a digraph of circumference l. Then the directed treewidth, DAG-width and Kelly-width of G are at most l + 1.
The above theorem can be seen as a mini mini mini directed grid minor theorem. I will be using this theorem in future papers to make progress towards a directed grid minor theorem. Stay tuned !!
# Open problems for 2014
Wish you all a Very Happy New Year. Here is a list of my 10 favorite open problems for 2014. They belong to several research areas inside discrete mathematics and theoretical computer science. Some of them are baby steps towards resolving much bigger open problems. May this new year shed new light on these open problems.
• 2. Optimization : Improve the approximation factor for the undirected graphic TSP. The best known bound is 7/5 by Sebo and Vygen.
• 3. Algorithms : Prove that the tree-width of a planar graph can be computed in polynomial time (or) is NP-complete.
• 4. Fixed-parameter tractability : Treewidth and Pathwidth are known to be fixed-parameter tractable. Are directed treewidth/DAG-width/Kelly-width (generalizations of treewidth) and directed pathwidth (a generalization of pathwidth) fixed-parameter tractable ? This is a very important problem to understand the algorithmic and structural differences between undirected and directed width parameters.
• 5. Space complexity : Is Planar ST-connectvity in logspace ? This is perhaps the most natural special case of the NL vs L problem. Planar ST-connectivity is known to be in $UL \cap coUL$. Recently, Imai, Nakagawa, Pavan, Vinodchandran and Watanabe proved that it can be solved simultaneously in polynomial time and approximately O(√n) space.
• 6. Metric embedding : Is the minor-free embedding conjecture true for partial 3-trees (graphs of treewidth 3) ? Minor-free conjecture states that “every minor-free graph can be embedded in $l_1$ with constant distortion. The special case of planar graphs also seems very difficult. I think the special case of partial 3-trees is a very interesting baby step.
• 7. Structural graph theory : Characterize pfaffians of tree-width at most 3 (i.e., partial 3-trees). It is a long-standing open problem to give a nice characterization of pfaffians and design a polynomial time algorithm to decide if an input graph is a pfaffian. The special of partial 3-trees is an interesting baby step.
• 8. Structural graph theory : Prove that every minimal brick has at least four vertices of degree three. Bricks and braces are defined to better understand pfaffians. The characterization of pfaffian braces is known (more generally characterization of bipartite pfaffians is known). To understand pfaffians, it is important to understand the structure of bricks. Norine,Thomas proved that every minimal brick has at least three vertices of degree three and conjectured that every minimal brick has at least cn vertices of degree three.
• 9. Communication Complexity : Improve bounds for the log-rank conjecture. The best known bound is $O(\sqrt{rank})$
• 10. Approximation algorithms : Improve the approximation factor for the uniform sparsest cut problem. The best known factor is $O(\sqrt{logn})$.
Here are my conjectures for 2014 :)
• Weak Conjecture : at least one of the above 10 problems will be resolved in 2014.
• Conjecture : at least five of the above 10 problems will be resolved in 2014.
• Strong Conjecture : All of the above 10 problems will be resolved in 2014.
Have fun !!
# Forbidden Directed Minors and Kelly-width
Today’s post is about the following paper, a joint work with Qiuyi Zhang, one of my advisees. Qiuyi Zhang is an undergraduate (rising senior) in our mathematics department.
• Shiva Kintali, Qiuyi Zhang. Forbidden Directed Minors and Kelly-width. (Preprint available on my publications page)
It is well-known that an undirected graph is a partial 1-tree (i.e., a forest) if and only if it has no $K_3$ minor. We generalized this characterization to partial 1-DAGs. We proved that partial 1-DAGs are characterized by three forbidden directed minors, $K_3, N_4$ and $M_5$, shown in the following figure. We named the last two graphs as $N_4$ and $M_5$ because their bidirected edges resemble the letters N and M.
Partial k-trees characterize bounded treewidth graphs. Similarly, partial k-DAGs characterize bounded Kelly-width digraphs. Kelly-width is the best known generalization of treewidth to digraphs.
As mentioned in the paper, I have a series of upcoming papers (called Directed Minors) making progress towards a directed graph minor theorem (i.e., all digraphs are well-quasi-ordered by the directed minor relation). For more details of the directed minor relation, read the current paper. I will post about the upcoming results as and when the preprints are ready.
# TrueShelf 1.0
One year back (on 6/6/12) I announced a beta version of TrueShelf, a social-network for sharing exercises and puzzles especially in mathematics and computer science. After an year of testing and adding new features, now I can say that TrueShelf is out of beta.
TrueShelf turned out to be a very useful website. When students ask me for practice problems (or books) on a particular topic, I simply point them to trueshelf and tell them the tags related to that topic. When I am advising students on research projects, I first tell them to solve all related problems (in the first couple of weeks) to prepare them to read research papers.
Here are the features in TrueShelf 1.0.
• Post an exercise (or) multiple-choice question (or) video (or) notes.
• Solve any multiple-choice question directly on the website.
• Add topic and tags to any post
• Show text-books related to a post
• Show related posts for every post.
• View printable version (or) LaTex version of any post.
• Email / Tweet / share on facebook (or) Google+ any post directly from the post.
• Like (a.k.a upvote) any post.
Feel free to explore TrueShelf, contribute new exercises and let me know if you have any feedback (or) new features you want to see. You can also follow TrueShelf on facebooktwitter and google+. Here is a screenshot highlighting the important features.
# Graceful Diameter-6 Trees
Graceful Tree Conjecture is one of my favorite open problems (See this earlier post). Trees with diameter 4 and 5 are known to be graceful a decade ago.
One of my advisees, Matt Superdock, made progress towards proving that all diameter 6 trees are graceful. Matt is an undergraduate senior in our Mathematics Department. He proved that an interesting class of diameter 6 trees are graceful. His thesis is available here.
I hope his work motivates more researchers to make progress towards resolving Graceful Tree Conjecture, particularly for diameter 6 trees. Matt’s work really tests the limit of current techniques. Perhaps we need new techniques/insights to prove that all diameter 6 trees are graceful.
# Happy Birthday Paul Erdos
Today (March 26 2013) is the 100th Birthday of Paul Erdos. The title of my Blog is inspired by one of his famous sayings “My Brain is Open”. In one of my earlier posts I mentioned a book titled “The Man Who Loved Only Numbers” about his biography.
Paul Erdos published more than 1500 papers. Most of them left a legacy of open problems and conjectures. What is your favorite open problem from Erdos’s papers ? Leave a comment. Hope we can solve some of his open problems during this special year.
Here are some interesting links : | 2,140 | 9,292 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 9, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2014-52 | latest | en | 0.918962 |
https://codeproject.freetls.fastly.net/Answers/617862/Formula-to-calculate-a-number-between-0-and-8?cmt=475087 | 1,653,601,832,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662625600.87/warc/CC-MAIN-20220526193923-20220526223923-00684.warc.gz | 225,470,676 | 31,954 | 15,312,353 members
1.00/5 (1 vote)
See more:
Hello Codeproject,
Assuming I have a row that follows like so:
0 1 2
3 4 5
6 7 8
And the number starts at 4, but I can only add or remove one, so saying;
-1 and -1 would equal to 0.
-1 and 0 would equal to 3.
Any ideas?
Posted
Comments
[no name] 8-Jul-13 15:59pm
Smells of homework. What have you tried?
Deviant Sapphire 8-Jul-13 16:00pm
Not really homework, this is for a project of mine but I do not have an idea. I thought about multiplying the number by three, but that didn't work out well.
Solution 2
I think I know what you're trying to say here, so here's an attempt at solving this (if I'm way off let me know, your description is a little vague):
Well you could use a 2-dimensional array, representing your grid, then use your -1/0/+1 as a change in the indexes, but that's more than you really need.
Look at it this way, what changes do those numbers really result in? The first moves you left or right, no matter where you are, the results in adding the same to the number (e.g. you start at 4, the first number is +1, you move to 5, which is 4 + 1). So just add the first number to 4. The second number moves you up or down. These all result in a change of 3*[second number] (e.g. you;re at 3, the second number is -1, you add -1*3 to 3 and get: 3 + (-1 * 3) = 0).
So the simplest way is:
4 + [first number] + [second number] * 3
Comments
Deviant Sapphire 8-Jul-13 16:21pm
Thank you I really appreciate the help and this did solve it!
Tim Carmichael 8-Jul-13 16:44pm
Looking at the set of numbers and the offered solution, please keep in mind what the desired behaviour is if the starting point is already on a edge (ie: not 4).
If you start with 0, what do you want to happen if the first number is -1? Do you wrap to 2 or ignore the value?
Deviant Sapphire 9-Jul-13 21:04pm
No this is exactly what I meant, I just followed your logic and it's easier for me to figure out next problems. Thanks a lot! :)
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https://www.crazyengineers.com/groups/43/crack-the-number-lock-code.741 | 1,556,257,670,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578760477.95/warc/CC-MAIN-20190426053538-20190426075538-00359.warc.gz | 644,420,140 | 10,253 | Tough Puzzles & Brain Teasers
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Aswini Vi
Computer Science
22 Jan 2019
# Crack the number lock code.
A number lock had 3 digit key.Hints for key are listed below.
>> 6,8,2 - One number is correct and well placed.
>>6,4,5 - One number is correct but wrong place.
>>2,0,6 - Two numbers are correct but wrong places.
>>7,3,8 - Nothing is correct.
>>7,8,0 - One number is correct but wrong placed.
3mos ago
3mos ago
0 5 2
From statement 1&4, one of 2 and 6 is right. From statement 1&2, 2 is right and is in the third place. 6 is wrong.
From statement 3, 0&2 must be correct and 0 must be the first.
From statement 2, 4 cannot be right. So 5 must be right and it should be the middle number.
2mos ago
0 5 2
Aswini Vi
Computer Science
2mos ago
2mos ago
Ans : 0 5 2
From 1,3 lines 2/6 is correct.
From (4) 7,3,8 are not in code
so From (5) 0 is correct but not in correct place.
from (2) 5 is correct and not in correct place.
So by arranging we get 0 5 2 | 352 | 1,090 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2019-18 | latest | en | 0.898134 |
http://cboard.cprogramming.com/c-programming/89660-guessing-game-printable-thread.html | 1,477,267,718,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719453.9/warc/CC-MAIN-20161020183839-00316-ip-10-171-6-4.ec2.internal.warc.gz | 43,939,344 | 4,883 | # guessing game
• 05-08-2007
Sal79
guessing game
I am very new at C , Any help would be greatly appreciated
This function is made to check the 2d array if guess matches num and loop untill the user enters -1.
the function call is not the problem its the return.
Code:
```int Find(int a[][SIZE],int num) { int y; int x; int guess=0; while(num != -1) { for(x=0;x<SIZE;x++)/*loop 1*/ { for(y=0;y<SIZE;y++)/*loop 2*/ { guess = a[x][y]; if (num == guess ) { return 1; break; } else { printf("Keep Guessing:"); scanf("%d",&num); } }/* close loop 2*/ }/* close loop 1 */ }/*while */ return 0; }/*Find*/```
• 05-08-2007
first, two things i would suggest: make your loop a do..while rather then a while loop. the break statement after the return 1 will never get executed so remove it.
Quote:
the function call is not the problem its the return.
without knowing the problem, where do i start looking? what is the situation with the return?
• 05-08-2007
Sal79
it just keeps going diplay keep guessing over and over endless.
how could i break out of the while if the guess is right.
• 05-08-2007
can you post your full code so we can see how its being called and what variables are what when passed.
i was trying to understand what the parameters were... the function is to find a number in a 2-dimensional array right? with the 2nd parameter being an initial guess? also this is an A x A (square) array right, so why not use a[SIZE][SIZE] in the parameter list?
• 05-08-2007
Sal79
Code:
``` #define SIZE 4 #include <stdio.h> #include <stdlib.h> #include <time.h> int main() { int table[SIZE][SIZE]; int Find(int a[][SIZE],int); int num; int find; int x,y; srand(time(NULL)); for(x=0;x<SIZE;x++) /* loop 1 */ { for(y=0;y<SIZE;y++) /* loop 2 */ { table[x][y]=100+rand()%101; } /* close loop 2 */ } /* close loop 1 */ printf("enter a number\n"); scanf("%d",&num); find = Find(table,num); if(find==1) { printf("good job you win"); } else { printf("sorry you lost"); } } int Find(int a[][SIZE],int num) { int y; int x; int guess=0; while(num != -1) { for(x=0;x<SIZE;x++)/*loop 1*/ { for(y=0;y<SIZE;y++)/*loop 2*/ { guess = a[x][y]; if (num == guess ) { return 1; break; } else { printf("Keep Guessing:"); scanf("%d",&num); } }/* close loop 2*/ }/* close loop 1 */ }/*while */ return 0; }/*Find*/```
this is the rest of my code
• 05-08-2007
thats not the rest of your code.. if it is, then you cant get it to even compile (which you said you can, therefore this isnt all of your code).
please just open your file, select all, and copy then paste here.
• 05-08-2007
Sal79
Sorry i had copied the wrong file
it just loops endlessly
• 05-08-2007
ulillillia
All I can think of is to add some temporary printf statements to check to see if it's getting to some parts of your loop. As a side note, a break in an inner loop will not cause the outer loop to also break, but seeing a "return" in there, it would just terminate the whole function whether or not the loops' conditions are true or false.
Better indentation can also be helpful.
• 05-08-2007
im still waiting for the answers from post #4 in this thread.
if the function is supposed to search through the entire array for the number, it isnt working. for the first iteration it compares the number given by the user with table[0][0]. if they arent equal it will read another number, and compare that to table[0][1], and so on. is this what you want?
• 05-08-2007
Sal79
It should? go through table[0][0] table[0][1] table[0][2] table[0][3]
table[1][0] table[1][1] table[1][2] table[1][3]
table[2][0] table[2][1] table[2][2] table[2][3]
table[3][0] table[3][1] table[3][2] table[3][3]
and check to see if num = said table in the nested for loop.
• 05-08-2007
if you want to guess a number and check the entire array for at least one occurance of it, you need to rethink your loops and move stuff around.. go through on paper what is happening in your Find function and you will see your simple mistake.
• 05-08-2007
Sal79
Code:
```int Find(int a[][SIZE],int num) { int y; int x; int guess=0; do { for(x=0;x<SIZE;x++)/*loop 1*/ { for(y=0;y<SIZE;y++)/*loop 2*/ { guess = a[x][y]; printf("%d",a[x+1][y]); if (num != guess ) { printf("Keep on Guessing:"); scanf("%d",&num); } else { return 1; } }/* close loop 2*/ }/* close loop 1 */ }while (num!=guess||num!=-1);/*while */ return 0; }/*Find*/```
this is some things i have changed. I hope i just didn't make something simple much harder
• 05-08-2007
```int Find(int a[][SIZE],int num) { int y; int x; int guess=0; while (1) { printf("Guess a number:"); scanf("%d",&num); if (num == -1) return 0; for(x=0;x<SIZE;x++)/*loop 1*/ { for(y=0;y<SIZE;y++)/*loop 2*/ { guess = a[x][y]; if (num == guess ) return 1; }/* close loop 2*/ }/* close loop 1 */ }/*do */ }/*Find*/``` | 1,645 | 5,924 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2016-44 | latest | en | 0.548023 |
http://www.convertaz.com/convert-276-milliliter-to-l/ | 1,568,892,958,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573476.67/warc/CC-MAIN-20190919101533-20190919123533-00157.warc.gz | 229,226,111 | 17,139 | # Convert 276 milliliter to l
## Conversion details
To convert milliliter to l use the following formula:
1 milliliter equals 0.001 l
So, to convert 276 milliliter to l, multiply 0.001 by 276 i.e.,
276 milliliter = 0.001 * 276 l = 0.276 l
For conversion tables, definitions and more information on the milliliter and l units scroll down or use the related milliliter and l quick access menus located at the top left side of the page.
### From 100.00 to 4000.00 milliliters, 40 entries
100 milliliters = 0.1 l
200 milliliters = 0.2 l
300 milliliters = 0.3 l
400 milliliters = 0.4 l
500 milliliters = 0.5 l
600 milliliters = 0.6 l
700 milliliters = 0.7 l
800 milliliters = 0.8 l
900 milliliters = 1 l
1000 milliliters = 1 l
1100 milliliters = 1.1 litres
1200 milliliters = 1.2 litres
1300 milliliters = 1.3 litres
1400 milliliters = 1.4 litres
1500 milliliters = 1.5 litres
1600 milliliters = 1.6 litres
1700 milliliters = 1.7 litres
1800 milliliters = 1.8 litres
1900 milliliters = 2 litres
2000 milliliters = 2 litres
2100 milliliters = 2.1 litres
2200 milliliters = 2.2 litres
2300 milliliters = 2.3 litres
2400 milliliters = 2.4 litres
2500 milliliters = 2.5 litres
2600 milliliters = 2.6 litres
2700 milliliters = 2.7 litres
2800 milliliters = 2.8 litres
2900 milliliters = 3 litres
3000 milliliters = 3 litres
3100 milliliters = 3.1 litres
3200 milliliters = 3.2 litres
3300 milliliters = 3.3 litres
3400 milliliters = 3.4 litres
3500 milliliters = 3.5 litres
3600 milliliters = 3.6 litres
3700 milliliters = 3.7 litres
3800 milliliters = 3.8 litres
3900 milliliters = 4 litres
4000 milliliters = 4 litres
Click here for a list of all conversion tables of milliliter to other compatible units.
## milliliter
Milliliter is a subdivision of the liter unit. The milli prefix stands for 0.001 therefore, 1 milliliter = 0.001 liter units. Liter is a unit of measurement of volume. The definition for liter is the following:
1 milliliter is equal to 1/1000000th of a liter.
The symbol for milliliter is ml
## litre
Litre is a unit of measurement of volume. The definition for litre is the following:
A litre is equal to 1 cubic decimeter or 0.001 cubic meters.
The symbol for litre is L
## Other people are also searching for information on milliliter conversions.
Following are the most recent questions containing milliliter. Click on a link to see the corresponding answer.
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Home | Base units | Units | Conversion tables | Unit conversion calculator | 874 | 2,690 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2019-39 | longest | en | 0.550612 |
https://www.askiitians.com/forums/Wave-Motion/what-is-the-difference-between-transverse-and-long_235635.htm | 1,721,099,642,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514726.17/warc/CC-MAIN-20240716015512-20240716045512-00414.warc.gz | 572,734,627 | 43,338 | # What is the difference between transverse and longitudinal waves and what is the relation between motion of wave and compression, rarefraction.????
Arun
25750 Points
5 years ago
Transverse Waves: Displacement of the medium is perpendicular to the direction of propagation of the wave.
To understand this it is good to think of a rope being held still by person B and being moved up and down by person A. The direction of propagation is from person A to B, so you will see the waves move along this way. But the displacement will be up and down.
Can travel in solids, but not in liquids and gas.
Longitudinal Waves: Displacement of the medium is parallel to the direction of propagation of the wave.
A good example for this is a slinky being pushed along the table, the propagation will be along the table and so will the displacement of all the 'rings'.
Can travel through all states of matter.
eg. Sound waves
longitudinal waves have compressions and rarefactions. Compression. A compression is a region in a longitudinal wave where the particles are closest together. Rarefaction. A rarefaction is a region in a longitudinal wave where the particles are furthest apart. | 278 | 1,182 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-30 | latest | en | 0.843868 |
http://numpy.d2l.ai/chapter_preliminaries/autograd.html | 1,571,123,239,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986657586.16/warc/CC-MAIN-20191015055525-20191015083025-00152.warc.gz | 139,700,098 | 13,308 | # 2.5. Automatic Differentiation¶
In machine learning, we train models, updating them successively so that they get better and better as they see more and more data. Usually, getting better means minimizing a loss function, a score that answers the question “how bad is our model?” This question is more subtle than it appears. Ultimately, what we really care about is producing a model that performs well on data that we have never seen before. But we can only fit the model to data that we can actually see. Thus we can decompose the task of fitting models into two key concerns: optimization the process of fitting our models to observed data and generalization the mathematical principles and practitioners wisdom that guide as to how to produce models whose validity extends beyond the exact set of datapoints used to train it.
This section addresses the calculation of derivatives, a crucial step in nearly all deep learning optimization algorithms. With neural networks, we typically choose loss functions that are differentiable with respect to our model’s parameters. Put simply, this means that for each parameter, we can determine how rapidly the loss would increase or decrease, were we to increase or decrease that parameter by an infinitessimally small amount. While the calculations for taking these derivatives are straightforward, requiring only some basic calculus, for complex models, working out the updates by hand can be a pain (and often error-prone).
The autograd package expedites this work by automatically calculating derivatives. And while many other libraries require that we compile a symbolic graph to take automatic derivatives, autograd allows us to take derivatives while writing ordinary imperative code. Every time we pass data through our model, autograd builds a graph on the fly, tracking which data combined through which operations to produce the output. This graph enables autograd to subsequently backpropagate gradients on command. Here, backpropagate simply means to trace through the compute graph, filling in the partial derivatives with respect to each parameter. If you are unfamiliar with some of the math, e.g., gradients, please refer to Section 16.2.
from mxnet import autograd, np, npx
npx.set_np()
## 2.5.1. A Simple Example¶
As a toy example, say that we are interested in differentiating the mapping $$y = 2\mathbf{x}^{\top}\mathbf{x}$$ with respect to the column vector $$\mathbf{x}$$. To start, let’s create the variable x and assign it an initial value.
x = np.arange(4)
x
array([0., 1., 2., 3.])
Note that before we even calculate the gradient of y with respect to x, we will need a place to store it. It’s important that we do not allocate new memory every time we take a derivative with respect to a parameter because we will often update the same parameters thousands or millions of times and could quickly run out of memory.
Note also that a gradient with respect to a vector $$x$$ is itself vector-valued and has the same shape as $$x$$. Thus it is intuitive that in code, we will access a gradient taken with respect to x as an attribute the ndarray x itself. We allocate memory for an ndarray’s gradient by invoking its attach_grad() method.
x.attach_grad()
After we calculate a gradient taken with respect to x, we will be able to access it via the .grad attribute. As a safe default, x.grad initializes as an array containing all zeros. That’s sensible because our most common use case for taking gradient in deep learning is to subsequently update parameters by adding (or subtracting) the gradient to maximize (or minimize) the differentiated function. By initializing the gradient to $$\mathbf{0}$$, we ensure that any update accidentally exectuted before a gradient has actually been calculated will not alter the variable’s value.
x.grad
array([0., 0., 0., 0.])
Now let’s calculate y. Because we wish to subsequently calculate gradients we want MXNet to generate a computation graph on the fly. We could imagine that MXNet would be turning on a recording device to capture the exact path by which each variable is generated.
Note that building the computation graph requires a nontrivial amount of computation. So MXNet will only build the graph when explicitly told to do so. We can invoke this behavior by placing our code inside a with autograd.record(): block.
with autograd.record():
y = 2.0 * np.dot(x, x)
y
array(28.)
Since x is an ndarray of length 4, np.dot will perform an inner product of x and x, yielding the scalar output that we assign to y. Next, we can automatically calculate the gradient of y with respect to each component of x by calling y’s backward function.
y.backward()
If we recheck the value of x.grad, we will find its contents overwritten by the newly calculated gradient.
x.grad
array([ 0., 4., 8., 12.])
The gradient of the function $$y = 2\mathbf{x}^{\top}\mathbf{x}$$ with respect to $$\mathbf{x}$$ should be $$4\mathbf{x}$$. Let’s quickly verify that our desired gradient was calculated correctly. If the two ndarrays are indeed the same, then their difference should consist of all zeros.
x.grad - 4 * x
array([0., 0., 0., 0.])
If we subsequently compute the gradient of another variable whose value was calculated as a function of x, the contents of x.grad will be overwritten.
with autograd.record():
y = x.sum()
y.backward()
array([1., 1., 1., 1.])
## 2.5.2. Backward for Non-scalar Variable¶
Technically, when y is not a scalar, the most natural interpretation of the gradient of y (a vector of length $$m$$) with respect to x (a vector of length $$n$$) is the Jacobian (an $$m\times n$$ matrix). For higher-order and higher-dimensional $$y$$ and $$x$$, the Jacobian could be a gnarly high order tensor and complex to compute (refer to Section 16.2).
However, while these more exotic objects do show up in advanced machine learning (including in deep learning), more often when we are calling backward on a vector, we are trying to calculate the derivatives of the loss functions for each constitutent of a batch of training examples. Here, our intent is not to calculate the Jacobian but rather the sum of the partial derivatives computed individuall for each example in the batch.
Thus when we invoke backwards on a vector-valued variable, MXNet assumes that we want the sum of the gradients. In short, MXNet, will create a new scalar variable by summing the elements in y, and compute the gradient of that variable with respect to x.
with autograd.record(): # y is a vector
y = x * x
y.backward()
u = x.copy()
with autograd.record(): # v is scalar
v = (u * u).sum()
v.backward()
array([0., 0., 0., 0.])
Already you know enough to employ autograd and ndarray successfully to develop many practical models. While the rest of this section is not necessary just yet, we touch on a few advanced topics for completeness.
### 2.5.3.1. Detach Computations¶
Sometimes, we wish to move some calculations outside of the recorded computation graph. For example, say that y was calculated as a function of x. And that subsequently z was calcatated a function of both y and x. Now, imagine that we wanted to calculate the gradient of z with respect to x, but wanted for some reason to treat y as a constant, and only take into account the role that x played after y was calculated.
Here, we can call u = y.detach() to return a new variable that has the same values as y but discards any information about how u was computed. In other words, the gradient will not flow backwards through u to x. This will provide the same functionality as if we had calculated u as a function of x outside of the autograd.record scope, yielding a u that will be treated as a constant in any called to backward. The following backward computes $$\partial (u \odot x)/\partial x$$ instead of $$\partial (x \odot x \odot x) /\partial x$$, where $$\odot$$ stands for elementwise multiplication.
with autograd.record():
y = x * x
u = y.detach()
z = u * x
z.backward()
array([0., 0., 0., 0.])
Since the computation of $$y$$ was recorded, we can subsequently call y.backward() to get $$\partial y/\partial x = 2x$$.
y.backward()
array([0., 0., 0., 0.])
## 2.5.4. Attach Gradients to Internal Variables¶
Attaching gradients to a variable x implicitly calls x=x.detach(). If x is computed based on other variables, this part of computation will not be used in the backward function.
y = np.ones(4) * 2
u = x * y
u.attach_grad() # implicitly run u = u.detach()
z = u + x
z.backward()
[1. 1. 1. 1.]
[1. 1. 1. 1.]
[0. 0. 0. 0.]
Detaching allows to breaks the computation into several parts. We could use chain rule Section 16.2 to compute the gradient for the whole computation. Assume $$u = f(x)$$ and $$z = g(u)$$, by chain rule we have $$\frac{dz}{dx} = \frac{dz}{du} \frac{du}{dx}.$$ To compute $$\frac{dz}{du}$$, we can first detach $$u$$ from the computation and then call z.backward() to compute the first term.
y = np.ones(4) * 2
u = x * y
v = u.detach() # u still keeps the computation graph
z = v + x
z.backward()
[1. 1. 1. 1.]
[0. 0. 0. 0.]
Subsequently, we can call u.backward() to compute the second term, but pass the first term as the head gradients to multiply both terms so that x.grad will contains $$\frac{dz}{dx}$$ instead of $$\frac{du}{dx}$$.
u.backward(v.grad)
[2. 2. 2. 2.]
[0. 1. 2. 3.]
## 2.5.6. Computing the Gradient of Python Control Flow¶
One benefit of using automatic differentiation is that even if building the computational graph of a function required passing through a maze of Python control flow (e.g. conditionals, loops, and arbitrary function calls), we can still calculate the gradient of the resulting variable. In the following snippet, note that the number of iterations of the while loop and the evaluation of the if statement both depend on the value of the input b.
def f(a):
b = a * 2
while np.abs(b).sum() < 1000:
b = b * 2
if b.sum() > 0:
c = b
else:
c = 100 * b
return c
Again to compute gradients, we just need to record the calculation and then call the backward function.
a = np.random.normal()
d = f(a)
d.backward()
We can now analyze the f function defined above. Note that it is piecewise linear in its input a. In other words, for any a there exists some constant such that for a given range f(a) = g * a. Consequently d / a allows us to verify that the gradient is correct:
print(a.grad == (d / a))
1.0
## 2.5.7. Training Mode and Prediction Mode¶
As we have seen, after we call autograd.record, MXNet logs the operations in the following block. There is one more subtle detail to be aware of. Additionally, autograd.record will change the running mode from prediction mode to training mode. We can verify this behavior by calling the is_training function.
print(autograd.is_training())
False
True
When we get to complicated deep learning models, we will encounter some algorithms where the model behaves differently during training and when we subsequently use it to make predictions. The popular neural network techniques dropout Section 4.6 and batch normalization Section 7.5 both exhibit this characteristic. In other cases, our models may store auxiliary variables in training mode for purposes of make computing gradients easier that are not necessary at prediction time. We will cover these differences in detail in later chapters.
## 2.5.8. Summary¶
• MXNet provides an autograd package to automate the calculation of derivatives. To use it, we first attach gradients to those variables with respect to which we desire partial derivartives. We then record the computation of our target value, executed its backward function, and access the resulting gradient via our variable’s grad attribute.
• We can detach gradients and pass head gradients to the backward function to control the part of the computation will be used in the backward function.
• The running modes of MXNet include training mode and prediction mode. We can determine the running mode by calling autograd.is_training().
## 2.5.9. Exercises¶
1. Try to run y.backward() twice.
2. In the control flow example where we calculate the derivative of d with respect to a, what would happen if we changed the variable a to a random vector or matrix. At this point, the result of the calculation f(a) is no longer a scalar. What happens to the result? How do we analyze this?
3. Redesign an example of finding the gradient of the control flow. Run and analyze the result.
4. In a second-price auction (such as in eBay or in computational advertising), the winning bidder pays the second-highest price. Compute the gradient of the final price with respect to the winning bidder’s bid using autograd. What does the result tell you about the mechanism? If you are curious to learn more about second-price auctions, check out this paper by Edelman, Ostrovski and Schwartz, 2005.
5. Why is the second derivative much more expensive to compute than the first derivative?
6. Derive the head gradient relationship for the chain rule. If you get stuck, use the “Chain Rule” article on Wikipedia.
7. Assume $$f(x) = \sin(x)$$. Plot $$f(x)$$ and $$\frac{df(x)}{dx}$$ on a graph, where you computed the latter without any symbolic calculations, i.e. without exploiting that $$f'(x) = \cos(x)$$. | 3,137 | 13,280 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2019-43 | latest | en | 0.932292 |
http://www.independent.ie/business/farming/measuring-grass-using-a-simple-system-brings-major-advantages-26725072.html | 1,503,089,758,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105108.31/warc/CC-MAIN-20170818194744-20170818214744-00179.warc.gz | 600,612,190 | 25,396 | Friday 18 August 2017
Hi °C | Lo °C Change
# Measuring grass using a simple system brings major advantages
What a difference one year can make. This time last year we were still feeding nuts and fodder-beet to lambed ewes. The only ewes getting any food but grass this spring are the few triplets and hoggets with twins.
By the end of March our grass was almost all eaten. To boost growth, 1 1/4 bags of 24-2.2-4.5 were spread on any ground that was grazed. The response has been really good with growth of 45kg/day recorded.
I measured grass with a plate metre. Firstly, record the number on the plate metre (eg 11,029). Then take 10 measurements randomly throughout the paddock.
This could give a recorded number (eg 11,141). To do your sum, subtract the first number from the second number and divide the answer by 20 to get the height of the grass.
11,141 - 11,029 = 112 ÷ 20 = 5.6cm.
This is your overall grass height so you subtract what will not be eaten. We aim to leave 3.5cm behind so there is 2.1cm to eat. Each centimetrein height is 200kg/ha, which gives 420kg/ha. Multiply by the size of the paddock -- 1.5ha gives 630kg of grass.
5.6cm - 3.5cm = 2.1cm x 200kg x 1.5ha = 630kg
Now allow for what each ewe will need to eat. A ewe rearing twin lambs of four weeks old needs 3kg/day. So a group of 50 ewes will need 150kg/day. We divide our total cover that will be eaten by our daily requirement (630kg ÷ 150kg) to give us 4.2 days grass for the 50 ewes and their lambs. I hope to manage my grass by using this method for this year. I will know when I have too much grass and I can take out paddocks for baled silage, and when there's not enough I can adjust numbers, spread fertiliser or wean the lambs earlier.
The advantage of measuring grass for me is better quality, less fertiliser, knowing what grass is ahead of each group and allowing me to sell lambs earlier and leave more grass for ewes later in the year.
Dose
Lambs are doing very well on good grass and plenty of milk from their mothers. Our next big job will be giving the lambs their first dose.
This will be a levamisole-type drench at about 5-6 weeks and it is for nematodirus worms. I hope that this dose will be given the same day as the lambs will be weighed. This 40-day weight is recorded using an electronic weigher with the electronic tags in the lambs linking them to their birth weight and parentage.
The hoggets and repeat ewes are still lambing, with about 20 of each left. We've had no problems with the ewes but the one-year-old hoggets again suffered because no enzootic abortion vaccine has been available. From 94 in-lamb at scanning, we hope 78 rear lambs. This is very disappointing to have 16 without lambs, and means a loss of potential income in a year where lamb prices look good. If this vaccine is unavailable next year, I definitely will not lamb ewe-lambs.
These all lambed outside by day and in the shed at night. A small amount of meal was given when they came indoors in the evening. This worked out very well to train them to come to the shed. We had some big single lambs that needed some assistance and a few trips to the vet.
The one good thing was that they all looked after their lambs well, no running away after giving birth or not letting the lambs suck. The twins are now on good grass getting 0.5kg of meal and the lambs will get creep from next week.
The singles are on grass only. Both groups will be joined together in another few weeks and all the lambs will get creep fed, with the lambs weaned at 12 weeks to give their mothers more time to put on weight before mating again. Just on a personal note to someone who taught me something very important at lambing -- patience. Happy birthday Mam.
John Large is a sheep farmer at Gortnahoe, Thurles, Co Tipperary
Indo Farming | 974 | 3,817 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2017-34 | latest | en | 0.953885 |
https://complex-analysis.com/content/taylor_series.html | 1,638,946,023,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363445.41/warc/CC-MAIN-20211208053135-20211208083135-00047.warc.gz | 252,502,938 | 7,750 | # Taylor Series
## For Real Functions
Let $a\in \mathbb R$ and $f(x)$ be and infinitely differentiable function on an interval $I$ containing $a$. Then the one-dimensional Taylor series of $f$ around $a$ is given by
$f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+\cdots$
which can be written in the most compact form:
$f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n.$
Recall that, in real analysis, Taylor's theorem gives an approximation of a $k$-times differentiable function around a given point by a $k$-th order Taylor polynomial.
For example, the best linear approximation for $f(x)$ is $$f(x)\approx f(a)+f′(a)(x−a).$$ This linear approximation fits $f(x)$ with a line through $x=a$ that matches the slope of $f$ at $a$.
For a better approximation we can add other terms in the expansion. For instance, the best quadratic approximation is $$f(x)\approx f(a)+f'(a)(x−a)+\frac12 f''(a)(x−a)^2.$$
The following applet shows the partial sums of the Taylor series for a given function. Drag the slider to show more terms of the series. Drag the point a or change the function.
Sorry, the applet is not supported for small screens. Rotate your device to landscape. Or resize your window so it's more wide than tall.
## For Complex Functions
Suppose that a function $f$ is analytic throughout a disk $|z -z_0|< R$, centred at $z_0$ and with radius $R$. Then $f(z)$ has the power series representation \begin{eqnarray}\label{seriefunction} f(z)=\sum_{n=0}^{\infty} a_n(z-z_0)^n,\quad |z-z_0|<R, \end{eqnarray} where \begin{eqnarray} a_n=\frac{f^{(n)}(z_0)}{n!},\quad n=0,1,2,\ldots \end{eqnarray} That is, series (\ref{seriefunction}) converges to $f(z)$ when $z$ lies in the stated open disk.
Every complex power series (\ref{seriefunction}) has a radius of convergence. Analogous to the concept of an interval of convergence for real power series, a complex power series (\ref{seriefunction}) has a circle of convergence, which is the circle centered at $z_0$ of largest radius $R \gt 0$ for which (\ref{seriefunction}) converges at every point within the circle $|z−z_0|=R$. A power series converges absolutely at all points $z$ within its circle of convergence, that is, for all $z$ satisfying $|z − z_0| \lt R$, and diverges at all points $z$ exterior to the circle, that is, for all $z$ satisfying $|z−z_0| \gt R$. The radius of convergence can be:
1. $R = 0$ (in which case (\ref{seriefunction}) converges only at its center $z = z_0$),
2. $R$ a finite positive number (in which case (\ref{seriefunction}) converges at all interior points of the circle $|z − z_0| = R)$, or
3. $R = \infty$ (in which case (\ref{seriefunction}) converges for all $z$).
The radius of convergence can be calculated using the ratio test of convergence. For example, if:
1. $\displaystyle \lim_{n\rightarrow \infty} \left| \frac{a_{n+1}}{a_n}\right| = L\neq 0$, the radius of convergence is $R=\dfrac{1}{L}$;
2. $\displaystyle \lim_{n\rightarrow \infty} \left| \frac{a_{n+1}}{a_n}\right|= 0$, the radius of convergence is $R=\infty$;
3. $\displaystyle \lim_{n\rightarrow \infty} \left| \frac{a_{n+1}}{a_n}\right|= \infty$, the radius of convergence is $R=0$.
## Dynamic Exploration
Use the following applet to explore Taylor series representations and its radius of convergence which depends on the value of $z_0$.
On the left side of the applet below, a phase portrait of a complex function is displayed. On the right side, you can see the approximation of the function through it's Taylor polynomials at the blue base point $z_0$. The complex function, the base point $z_0$, the order of the polynomial (vertical slider) and the zoom (horizontal slider) can be modified.
f(z) =
You can also select one function of the following list:
Sorry, the applet is not supported for small screens. Rotate your device to landscape. Or resize your window so it's more wide than tall.
## Maclaurin series
A Taylor series with centre $z_0=0$ $f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}z^n$ is referred to as Maclaurin series.
Some important Maclaurin series are: \begin{eqnarray*} \displaystyle \frac{1}{1-z}&=& \sum_{n=0}^{\infty} z^n, \quad |z|\lt 1; \\ \displaystyle e^z &=& \sum_{n=0}^{\infty} \frac{z^n}{n!} \quad |z|\lt \infty;\\ \displaystyle \sin z &=& \sum_{n=0}^{\infty} (-1)^n\frac{z^{2n+1}}{n!} \quad |z|\lt \infty;\\ \displaystyle \cos z &=& \sum_{n=0}^{\infty} (-1)^n \frac{z^{2n}}{n!} \quad |z|\lt \infty;\\ \displaystyle \sinh z &=& \sum_{n=0}^{\infty} \frac{z^{2n+1}}{n!} \quad |z|\lt \infty;\\ \displaystyle \cosh z &=& \sum_{n=0}^{\infty} \frac{z^{2n}}{n!} \quad |z|\lt \infty; \end{eqnarray*}
Exercise: Find the Maclaurin series expansion of the function $f(z)=\frac{z}{z^4+9}$ and calculate the radius of convergence.
Note: The applet was originally written by Aaron Montag using CindyJS. The source can be found at GitHub.
NEXT: Laurent Series | 1,574 | 4,905 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2021-49 | longest | en | 0.77594 |
https://www.jiskha.com/similar?question=Short+answer+find+the+quotient+6x%5E3-x%5E2-7x-9+------------+2x%2B3&page=122 | 1,566,775,525,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027330907.46/warc/CC-MAIN-20190825215958-20190826001958-00391.warc.gz | 860,469,339 | 28,846 | # Short answer find the quotient 6x^3-x^2-7x-9 ------------ 2x+3
143,010 questions, page 122
1. ## biochem
For an enzyme that follows simple Michaelis–Menten kinetics, what is the value of Vmax if V0 is equal to 1 mmol minute−1 when [S] = 1/10 KM I keep getting 11 for the answer but the answer key in my book says 1.1 umol. what am i doing wrong?
asked by sean on October 15, 2018
2. ## college psych
Information to answer the question "What color is your vehicle? Would be located in blank according to the psychoanalytic theory. a) The Ego b) The conscious c) The unconscious d) The Superego e) The Preconscious I believe the answer is b not sure
asked by Michele on December 10, 2008
3. ## English
The sentence, I don't know where is it, represents an error in word order. 1. probably where often formed a question. 2. probably because a noun clause was not recognized My answer answer 1 & 2 are correct statements
asked by Jim Collum on June 10, 2013
4. ## Help?
I thought that when you asked questions on here they would try to help you figured out the answer, i know there not supposed to tell you the answer, but i mean why do some people just say look it up on google? Isn't that kind of defeating the purpose of
asked by No on November 18, 2008
5. ## Anatomy and Physiology
Hello! Thanks for checking this question! ____ 26. At approximately what age does nerve conduction achieve maximum capacity? (1 point) a) 5 b) 10 c) 20 d) 30 My Answer: A Could someone please check my answer? Thanks! - Da Fash
asked by Da Fash on November 9, 2017
6. ## Probability
If X is a normal random variable with mean 11, and if the probability that X is less than 12.10 is .72 , then what is the standard deviation of X? Select one: a. 3.00 b. 6.25 c. 2.50 d. 1.25 Please note that the answer for the above is stated as c.2.50.
asked by Alia on January 4, 2013
7. ## Math
tan(arccosx)=? I know the answer is supposed to be sqrt(1-x^2)/2, but how would I go about arriving at that answer? I've searched my textbook and Internet but still haven't got a clue.. Is it simply an identity that I must memorize? Please help
asked by Anonymous on September 15, 2011
8. ## chemistry
why are the flair lights at the air port yellow/orange instead of white? my teacher gave me until monday 18th to answer this and i think he expects a scientific answer but i have nothing
asked by vanessa on April 17, 2010
9. ## government
Please can you help me to answer these questions. 1. Does Britain rule itself today? I wasn't sure how to answer this since it joined the EEC in 2009. 2.Is a country ruled by another called dependent, or a colony? Many thanks for your help.
asked by Sam on March 9, 2010
10. ## Math
I need someone to check over this assignment that is for extra credit. I have include the problems as well as the answers . I thank you. Multiple Choice Questions (On Multiple Choice Questions, you do not need to show your work unless asked to.) 1. Write
asked by tamara jones on September 5, 2009
11. ## science
If the block has a mass 12.3 kg and slides a distance of 6.9 m down the inlcine, calculate the distance x the spring is compressed if the coefficient of kinetic friction between the block and the incline is 0.3 and the spring constant is 17.4 N/m. Input
asked by xolani on March 14, 2013
12. ## Inter. Algebra
Count Iblis what is the answer to 2 square root sign 18x? The instructions say to simplify or reduce. 2*(18x)1/2= 2*(9*2)1/2= 2*3(2)1/2= 6*(2)1/2 If you can leave it in this form, the above is the way to end. If you are to continue to a final answer, then,
asked by Andrea on December 2, 2006
13. ## Physics
While following the directions on a treasure map, a pirate walks 36.6 m north, then turns and walks 6.3 m east. A.) What is the magnitude of the single straight-line displacement that the pirate could have taken to reach the treasure? Answer in units of m.
asked by Luke on November 8, 2012
14. ## Spanish
Is this correct how I answered these two questions? Donde vive presidente de los Estados Unidos? My answer: El presidente vive en Washington, D.C. De donde es tu amigo? Answer: El amigo es de Colombia. I know I didn't use accent marks-I will put those in,
asked by Julia on September 23, 2010
15. ## Calculus
this homework is due in one hour and i have no idea how to answer these questions. 1. A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 20 ft/s^2. What is the distance covered before the car comes to a
asked by Sam on November 16, 2011
16. ## Math
Mr. Jenkins is going to buy museum tickets for a class trip. The museum director quotes him the following prices: • 20 tickets cost a total of \$250. • 30 tickets cost a total of \$375. • 50 tickets cost a total of \$625. a. Based on these prices, is
asked by Anonymous on January 14, 2015
17. ## French
Can you please check these Choose A. Le caissier m'a donné 200 euros en liquide. B. Je suis toujours fauché. C. Le caissier m'a donné 10 billets de dix. D. J'aime mettre de l'argent de côté. 1. -------- J'ai touché un chèque de 200 euros. Answer: A
asked by Vanessa on March 15, 2010
18. ## Grammar
Can someone please help me with these questions? I have noo idea about them! Identify the function of the relative pronoun in the adjective clause. 1.The game was designed by Zack, who studies computer programming. 2.We all enjoyed the play that you wrote.
asked by mysterychicken on May 3, 2009
19. ## math
10.On 16 april 1990, RM5000 was invested at 6% compounded semi annually. Find the amount accumulated on 16 october 1993 if the rate was charged to 8% compounded quarterly beginning 16 april 1992. the answer is : RM 6339.24 i want to know the step to solve
asked by fizz on April 24, 2014
20. ## statistics
In a particular hospital, newborn babies were delivered yesterday. Here are their weights (in ounces): 106 104 96 123 116 Assuming that these weights constitute an entire population, find the standard deviation of the population. Round your answer to at
asked by Nathan on April 19, 2010
21. ## Physics
"a vector r has a magnitude of 18 units and makes a 30 degree angle with respect to the x axis. find the vector components of r using a protractor and some graph paper to verify your answer by drawing r and measuring the length of the lines representing
asked by Alyssa on September 14, 2015
22. ## Geometry
Jason designed an arch made of wrought iron for the top of a mall entrance. The 11 segments between the two concentric circles are each 1.25 m long. Find the total length of wrought iron used to make the structure. Round the answer to the nearest meter.
asked by Sonya on January 14, 2013
23. ## physics
1100g of ice at -50C is mixed with 100g of steam at 120C. Find out how much ice melts? The implied final temperature is 0C. Cice=0.5 Csteam=0.46 Hv=539 Hf =80 here is my work 1100(0.5)(0+50)+80m = 100(0.46)(120-0)+540(100) i am not getting the right answer
asked by jeff on December 5, 2009
24. ## Algebra
Assume the car can be purchased for 0% down for 60 months (in lieu of rebate). A car with a sticker price of \$42,300 with factory and dealer rebates of \$5,100 (a) Find the monthly payment if financed for 60 months at 0% APR. (Round your answer to the
asked by Anonymous on June 23, 2013
25. ## Physics
An airplane flies 340 km due east, makes a right-angle turn, and then flies 390 km due north. Find the magnitude of the plane’s displace- ment from its starting point. Neglect the curvature of the earth. Answer in units of km
asked by Crystal on September 30, 2012
26. ## Science/Weather
Satellite images help meteorologists in all the following ways except: a) monitoring changes in the strength of storms b) locating areas of high and low pressure systems. c) predicting rainfall amounts d) recording daily high and low temperatures. I think
asked by HiHi12345Kate on February 13, 2017
27. ## calculus
I posted this below, and no one answered. Please help with this very complicated question! Using 3(x-3)(x^2-6x+23)^1/2, as the chain rule differentiation of f(x)=(x^2-6x+23). Please explain how I find the general solution to the differential equation
asked by Amy on April 19, 2011
28. ## PHYSICS
Given: The energy equivalent of one gallon of gasoline is 1.3 × 108 J. A compact car has a mass of 847 kg, and its efficiency is rated at 18.1%. (That is, 18.1% of the input power available is delivered to the wheels.) Find the amount of gasoline used to
asked by anonymous on April 7, 2011
29. ## calculus
Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle and label its height and width. y=tan(11x) and y= sin(11x). -pi/33>= x
asked by angel on February 12, 2010
30. ## social studies
Children whose parents encourage them to use gestures in their development of communication: A)ave difficulty mastering a large vocabulary. B)have better comprehension. C)are frustrated in their communication. D) watch people's gestures more than they
asked by missy on October 10, 2011
31. ## algebra
on an expressway, the recommended safe distance between cars in feet is given by 0.019v^2+v+14, where v is the speed of the car in miles per hour. Find the safe distance when the following takes place( round answer to three decimal places). v=35 mph
asked by Marc on September 27, 2014
32. ## Math: Calculus - Geometric Vectors
Find the magnitude and the direction of the resultant of each of the following systems of forces using geometric vectors. a) Forces of 3 N and 8 N acting at an angle of 60 degrees to each other. Please help me with this question. I don't understand the
asked by Anonymous on February 17, 2008
33. ## Physics
A subway train starts from rest at a station and accelerates at a rate of 1.50m/s^2 for 13.3s . It runs at constant speed for 71.0s and slows down at a rate of 3.48m/s^2 until it stops at the next station. Find the total distance covered. Express your
asked by Kim on January 17, 2012
34. ## Physics
A subway train starts from rest at a station and accelerates at a rate of 1.50m/s^2 for 13.3s . It runs at constant speed for 71.0s and slows down at a rate of 3.48m/s^2 until it stops at the next station. Find the total distance covered. Express your
asked by Kim on January 17, 2012
35. ## statistics
2. The new Twinkle bulb has a standard deviation hours. A random sample of 77 light bulbs is selected from inventory. The sample mean was found to be 492 hours. a. Find the margin of error E for a 90% confidence interval. Round your answer to the nearest
asked by stefanie on May 2, 2010
36. ## algebra
on an expressway, the recommended safe distance between cars in feet is given by 0.019v^2+v+14, where v is the speed of the car in miles per hour. Find the safe distance when the following takes place( round answer to three decimal places). v=35 mph
asked by Marc on September 27, 2014
37. ## physics
A 25 kg cannon ball is fired from a cannon with muzzle speed of 1520 m/s at an angle of 35.5 ◦ with the horizontal. The acceleration of gravity is 9.8 m/s 2 . Use conservation of mechanical energy to find the maximum height reached by first ball. Answer
asked by keegan on February 1, 2015
38. ## managerial economics
Where might I find the answer to the following: US cigarette makers face enormous punitve damage penalties after a series of class action lawsuits. In spite of these the cigarette makers were able to avoid bankruptcy. I need to explain why and how using
asked by Linda S on February 16, 2011
39. ## calculus
I posted this below, and no one answered. Please help with this very complicated question! Using 3(x-3)(x^2-6x+23)^1/2, as the chain rule differentiation of f(x)=(x^2-6x+23). Please explain how I find the general solution to the differential equation
asked by Amy on April 21, 2011
40. ## Statistics
In a random sample of 508 judges, it was found that 281 were introverts. Let p represent the proportion of all judges who are introverts. Find a point estimate for p. Round your answer to four decimal places. I am needing assitance in trying to figure out
asked by Jacob on June 17, 2010
41. ## physics
A small ball of mass 65 g is suspended from a string of length 53 cm and whirled in a circle lying in the horizontal plane.If the string makes an angle of 28◦ with the vertical, find the centripetal force experienced by the ball. The acceleration of
asked by John on February 26, 2015
42. ## math
nina designed an arch made of wrought iron for the top of a mall entrance. the 11 segmentsbetween the two concentric semicircles are each 3 ft long. find the total length of the wrought iron used to make this structure. round your answer to the nearest
asked by keyla on February 20, 2012
43. ## managerial economic
Where might I find the answer to the following: US cigarette makers face enormous punitve damage penalties after a series of class action lawsuits. In spite of these the cigarette makers were able to avoid bankruptcy. I need to explain why and how using
asked by Linda on February 15, 2011
44. ## Spanish
Could you please explain the transalation and how to answer for the first and check my answer for the second. Estás en el correo. ¿Qué hora es? I think this says I'm in the mail and what hour is it-but that doesn't make sense. Please explain how I
asked by JamieLynne on November 29, 2010
45. ## algebra
A poll was taken of 100 students at a commuter campus to find out how they got to campus. The results are below. 35 said they drove alone. 38 rode in a carpool. 33 rode public transportation. 5 used both carpools and public transportation. 4 used both a
asked by matt on November 19, 2011
46. ## English
08verF(27-31) I need a help.I give a examination 2days later. Identify the one underlined word or phrase that should be corrected. 1.By the time(A) we read the paper,the news are old(B),since we've already heard(C) on TV what's happening(D) in the world.
asked by Bayarbold on June 24, 2011
47. ## Physics for Engineers
A playground is on the flat roof of a city school, 5.2 m above the street below (see figure). The vertical wall of the building is h = 6.70 m high, forming a 1.5-m-high railing around the playground. A ball has fallen to the street below, and a passerby
asked by Lauryn on September 14, 2011
48. ## spanish SRA MCGUINN
are these correct Que hiciste ayer? my answer YO huhica moas tarea Adonde fuiste el sabado pasado? My answer Estaba en las peliculas Estuduante para tu clase de espanol anteayer? My answer Estudie para la clase espanola ayer -- is it espanola or espanol?
asked by sam on January 23, 2008
49. ## Chemistry-
The question I have to answer is: If two peaks of gas chromatogram are too close, what changes in experimental conditions (factors affecting separation) could you make to separate them more effectively? Answer: I thought the answer would be the length of
asked by Soojung on September 8, 2009
50. ## Math
Learning about quadratic functions and equations, and I am struggling. It's so easy for others to understand, but math is my weakness, would appreciate all the help you could give. Notes and stuff would be great, as long as I understand. Quadratic
asked by -Untamed- on November 24, 2011
1. acupuncture is an example of chinese A. cooking B. medicine
asked by keykat on March 3, 2016
52. ## Chemistry
If the solubility of a gas in water is 1.22 g/L at 2.75 atm, what is its solubility (in g/L) at 1.0 atm? A. 0.44 g/L B. 3.97 g/L C. 2.25 g/L D. 3.36 g/L I think the answer is C.. i did 2.75 x 1.0 = 2.75 / 1.22 = 2.25 but people say the answer is A... why?
asked by Alexa on May 22, 2014
53. ## physics
a) a driver of mass 55kg needs to jump to a height of 1.1m above her take off height, at which point her speed needs to be 0.5m/s^. How fast just she be travelling the instant her feet leave the diving board? So we know eg=mgh =9.8*55*1.1 I assumed we're
asked by alex on March 28, 2017
54. ## Latin
I need help with this assignment. My answers are included but don't feel confident. Can you help? Directions: Conjugate the following verbs in the tense given in Latin and English 1. gero, gerere, gessi, gestus – to wear (present) answer: gero I wear 2.
asked by Jessie on November 26, 2015
55. ## French
Could you please check these thanks. Part 1 Directions: Choose the right word with the right statement. Word Bank A. donner B. téléphoner C. mettre D. décrocher E. commposer 1. ------ le numéro Answer: E 2. ----- le téléphoner Answer: D 3. ------- un
asked by Kesha on February 9, 2010
56. ## physics
can somebody shows me how to get to the answer key-- a)=2.51s, b)=19.9m, and c)=11.8i-16.1j A 1.57-kg particle initially at rest and at the origin of an x-y coordinate system is subjected to a time-dependent force of F(t) = (7.00ti − 8.00j)N with t in
asked by joy on February 8, 2018
57. ## Calculus AB
Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x) = 3x(1/3) + 6x(4/3). You must justify your answer using an analysis of f '(x) and f "(x). My start to a solution: x-values: (-1/8) and (-1/4) I got -1/8 from
asked by Vikram on January 11, 2016
58. ## physics
A 24.9 g marble sliding to the right at 22.0 cm/s overtakes and collides elastically with a 13.2 g marble moving in the same direction at 11.8 cm/s. After the collision, the 13.2 g marble moves to the right at 25.0 cm/s. Find the velocity of the 24.9 g
asked by luma on December 19, 2015
59. ## math
a square has an area of 169 square centimeters. find the area of the largest circle that could be drawn inside the square. round your answer to the nearest square centimeter.
asked by ash on April 10, 2013
60. ## religious education
please could someone help me find a websie about a man called billy mccurrie. i need to answer 2 questions about him:- what events made him not believe in God and what events made him believe in God. thankyou for your time to help me
asked by Ryder on November 7, 2009
61. ## math
write a five digit number that fits each description use only the digits that the athlete are wearing use each digit only once in each answer find the greatest number less than 1.8 1. The digits ate 41836
asked by Robyn on December 11, 2013
62. ## Physics
The density of a solid is 130gcm^-3 at the temperature of 25 degree Centrigate.Find the Density at 150 degree Celcius if the linear expansion of the solid is 2.0*10^(-5)K^-1. I got the answer to be 129.03.
asked by Sirneij on April 14, 2015
63. ## math
write a five digit number that fits each description use only the digits that the athlete are wearing use each digit only once in each answer find the greatest number less than 1.8 1. The digits ate 41836
asked by Robyn on December 12, 2013
64. ## geography
Which hemisphere is larger, The Eastern or Western? Can you please give me a web site to look at for this answer. I find it to be that they are the same size, but in terms of landmass the Eastern hemisphere is larger. Is that correct?
asked by Reed on November 8, 2010
65. ## Physics
The density of a solid is 130gcm^-3 at the temperature of 25 degree Centrigate.Find the Density at 150 degree Celcius if the linear expansion of the solid is 2.0*10^(-5)K^-1. I got the answer to be 129.03.
asked by Sirneij on April 14, 2015
66. ## Physics
A 26.3 g marble sliding to the right at 21.0 cm/s overtakes and collides elastically with a 13.3 g marble moving in the same direction at 12.2 cm/s. After the collision, the 13.3 g marble moves to the right at 23.7 cm/s. Find the velocity of the 26.3 g
asked by Chris on January 26, 2015
67. ## satistics
A test is composed of six multiple choice questions where each question has 4 choices. If the answer choices for each question are equally likely, find the probability of answering more than 4 questions correctly.
asked by Anonymous on October 20, 2010
68. ## Math
Im just a little confused on how to answer this question. A ball bounced 64% of the height from which it was dropped. The bounce was 72 cm high. What is the height from which the ball was dropped. I think there has to be an algebraic equation of some sort,
asked by Erika on May 15, 2011
69. ## Physics
In the circuit shown, find the voltage Vx (in Volts) if V = 17.0 Volts, R1 = 13.0 k ohms, R2 = 4.0 k ohms, R3 = 13.0 k ohms and R4 = 21.0 k ohms. Answer is -6.5 I don't understand how they got it, can someone please explain it with steps? Thanks tinypic. c
asked by christina on February 11, 2014
70. ## math
Four balls are selected at random without replacement from an urn containing three white balls and five blue balls. Find the probability that two or three of the balls are white. (Round your answer to three decimal places.)
asked by Anonymous on February 27, 2011
71. ## maths - ratio
the scale on a map is 1:2500, a pond is represented on the map as a shape whose perimeter is 12cms. Find the actual perimeter of the pond in kilometres. I came up with the answer 0.3. Is this correct?
asked by savannah xx on September 16, 2008
72. ## physics
In the circuit shown, find the voltage Vx (in Volts) if V = 17.0 Volts, R1 = 13.0 k ohms, R2 = 4.0 k ohms, R3 = 13.0 k ohms and R4 = 21.0 k ohms. Answer is -6.5 I don't understand how they got it, can someone please explain it with steps? Thanks tinypic. c
asked by christina on February 11, 2014
73. ## Math
a hollow metal sphere has an internal radius of 20cm and an external radius of 30cm , given that the density of the metal is 7.8, find the mass of the sphere , expressing your answer in kg.
asked by Steph on July 4, 2016
74. ## math
A 12-foot tree casts a 22-foot shadow. Find the angle of elevation from the tip of the shadow to the top of the tree. Round the answer to the nearest tenth.
asked by celina on October 7, 2014
75. ## college
A test is composed of six multiple choice questions where each question has 4 choices. If the answer choices for each question are equally likely, find the probability of answering 3 OR 4 questions correctly.
asked by poe on June 17, 2010
76. ## Math
1. Use 3.14 for pie to estimate the area of a circle. The diameter is given. Round your answer to the nearest hundredth if necessary. (the diameter is 16.8) 2. Find the missing length. Round to the nearest tenth, if necessary. (7,19)
asked by Math on May 8, 2017
77. ## Math
Andy divided his 1/3 jar of paint equally between 2 projects. What fraction of a jar of paint did Andy use for each project?Explain how to draw a model to find the answer.
asked by Kylea on November 29, 2017
78. ## Alegebra
A recipe calls for 2 cups of flour to make 24 cookies. Write and solve a proportion to find how much flour is needed to make 54 cookies. My answer: 4.5 because 54 divided by 24 is 2.25 and 2 x 2.25 is 4.5
asked by Mopdopnop on December 23, 2016
79. ## mathematics
I am supposed to find the line integral over C of (x+y)ds where C is the line segment from (0,1,1) to (3,2,2)? So far I have 3ti + (1+t)j + (1+t)k as the parameterisation of C, and when I make an integral from that, I get 3(sqrt(11)). But I am uncertain of
asked by JuliaF on December 3, 2018
80. ## Algebra
Consider the following matrix. A = 7, -5 -4, 3 Choose the correct description of A Find A^−1 if it exists. Answer either Choice 1, or Choice 2: CHOICE 1: A is nonsingular; that is, it has an inverse. A^-1 = _________ or CHOICE 2 : A is singular; that is,
asked by Shelby on March 17, 2017
81. ## Trigonometry
Find the angular speed in radians per sec. of the second hand on a clock. How long does it take for the second hand to make one rotation? And isn't one rotation 2pi radians? Let me know what you got as an answer.
asked by James on July 3, 2007
82. ## precal
please help i can't find the answer for this question. i can only manipulate one side of the equation and the end result has to equal the other side. Problem 1. ((1 + sinx)/cosx) + (cosx/(1 + sinx)) = 2secx
asked by Aly on November 2, 2012
83. ## math
find area of cirle 3.9cm radius to nearest tenth? a=pie 3.14 x r squared 3.14 x 3.9(15.21) 3.14 x15.21 =47.75 nearest tenth =47.8cm squared is this the correct answer?
asked by joshuah on May 23, 2010
84. ## pre-algebra
Rewrite the equation so the variable is alone. Solve to find the value of the variable. 42=-2h Divide both sides by 2. Then change the signs of both sides. That will provide the answer. Possibly in 1 step: Divide by -ve 2
asked by Julia on March 15, 2007
85. ## stat
A test is composed of six multiple choice questions where each question has 4 choices. If the answer choices for each question are equally likely, find the probability of answering more than 4 questions correctly
asked by laura on June 13, 2010
86. ## Geometry
graph each set of lines to form a triangle. find the area and perimeter. y=2, x=5 , and y=x Answer is Area = 4.5 units Perimeter = 6 + square root of 2. I don't know how to graph this. What does y=x mean?
asked by Ethan on July 4, 2012
87. ## Science
A 0.44 kg mass at the end of a horizontal spring is displaced 1.5 m and released, then moves in SHM at the end of the spring of force constant 2 N/m. Find the potential energy of the system when the spring is stretched 0.9 m. The acceleration of gravity is
asked by Emma on January 22, 2016
88. ## PHYSICS!!!
A straight horizontal pipe with a diameter of 1.0 cm and a length of 46 m carries oil with a coefficient of viscosity of 0.11 N · s/m2. At the output of the pipe, the flow rate is 8.4 10-5 m3/s and the pressure is 1.0 atmosphere. Find the gauge pressure
asked by k on November 11, 2012
89. ## physics!
A (10.25) kg bowling ball is hung on a (11.50) m long rope. It is then pulled back until the rope makes an angle of (39.0)o with the vertical and released. Find the tension in the rope when the ball is at the lowest point. Give your answer in N and with 3
asked by Idali on March 14, 2017
90. ## Science
A 0.44 kg mass at the end of a horizontal spring is displaced 1.5 m and released, then moves in SHM at the end of the spring of force constant 2 N/m. Find the potential energy of the system when the spring is stretched 0.9 m. The acceleration of gravity is
asked by Emma on January 22, 2016
91. ## math
Can someone give me the formula to work this out, I am not sure. Find the value of t for the difference between two means based on an assumption of normality and this information about two samples. (Use sample 1 - sample 2. Give your answer correct to two
asked by tracy10 on June 23, 2013
92. ## Statistics
Electricity bills in a certain city have mean \$108.73. Assume the bills are normally distributed with standard deviation \$10.60. A sample of 47 bills was selected for an audit. Find the 46 percentile for the sample mean. Write only a number as your answer.
asked by Tommy on October 6, 2015
93. ## A Geometry ?
Given that two sides of a triangle are 20 and 66. Find the range of possible measures for the third side, s. a. 20
asked by Ali on June 18, 2006
94. ## physics
You are trying to find out how high you have to pitch a water balloon in order for it to burst when it hits the ground. You discover that the balloon bursts when you have pitched it to a height of 14 m. With what velocity did the balloon hit the ground?
asked by jj on October 2, 2011
95. ## physics
You are trying to find out how high you have to pitch a water balloon in order for it to burst when it hits the ground. You discover that the balloon bursts when you have pitched it to a height of 17 m. With what velocity did the balloon hit the ground?
asked by Jason on November 19, 2014
96. ## Math
To determine the number of deer in a game preserve, a conservationist catches 443, tags them and then lets them loose. Later, 528 deer are caught and 176 are tagged. How many deer are in the preserve? What formula do I use to find the answer to this? Thank
asked by Ashli on June 16, 2008
97. ## Math
A moving conveyor is built to rise 4 ft for each 5 ft of horizontal change. Suppose the conveyor runs between two floors in a factory. Find the length of the conveyor if the vertical distance between floors is 8 feet. (Round your answer to three decimal
asked by Anonymous on August 20, 2018
98. ## Chemistry
How do I find the answer to this?? Cu metal displaces Ag+(aq) from an aqueous solution. Which of the following is correct? Ag is easier to oxidize than Cu Ag is a better reducing agent than Cu Cu2+ is a better oxidizing agent than Ag+ Ag+ is easier to
asked by Clueless on January 18, 2018 | 8,001 | 28,549 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2019-35 | latest | en | 0.960821 |
https://gitlab.cosma.dur.ac.uk/swift/swiftsim/-/commit/436c80b6fa874c1aafee5bc960376cdfe508adac | 1,675,684,919,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500339.37/warc/CC-MAIN-20230206113934-20230206143934-00154.warc.gz | 299,692,336 | 62,079 | Commit 436c80b6 by Tom Theuns
parent 888a3fe5
... ... @@ -8,7 +8,8 @@ iplot = 1 ; if iplot = 1, make plot of E/Lz conservation, else, simply compare f @physunits indir = './' basefile = 'Disk-Patch-dynamic_' ;basefile = 'Disc-Patch-dynamic_' basefile = 'Disc-Patch_' ; set properties of potential uL = phys.pc ; unit of length ... ... @@ -16,18 +17,27 @@ uM = phys.msun ; unit of mass uV = 1d5 ; unit of velocity ; properties of patch surface_density = 10. surface_density = 100. ; surface density of all mass, which generates the gravitational potential scale_height = 100. z_disk = 200.; z_disk = 200. ; fgas = 0.1 ; gas fraction gamma = 5./3. ; derived units constG = 10.^(alog10(phys.g)+alog10(uM)-2d0*alog10(uV)-alog10(uL)) ; pcentre = [0.,0.,z_disk] * pc / uL utherm = !pi * constG * surface_density * scale_height / (gamma-1.) temp = (utherm*uV^2)*phys.m_h/phys.kb soundspeed = sqrt(gamma * (gamma-1.) * utherm) t_dyn = sqrt(scale_height / (constG * surface_density)) rho0 = fgas*(surface_density)/(2.*scale_height) print,' dynamical time = ',t_dyn,' = ',t_dyn*UL/uV/(1d6*phys.yr),' Myr' print,' thermal energy per unit mass = ',utherm print,' central density = ',rho0,' = ',rho0*uM/uL^3/m_h,' particles/cm^3' print,' central temperature = ',temp lambda = 2 * !pi * phys.G^1.5 * (scale_height*uL)^1.5 * (surface_density * uM/uL^2)^0.5 * phys.m_h^2 / (gamma-1) / fgas print,' lambda = ',lambda stop ; infile = indir + basefile + '*' spawn,'ls -1 '+infile,res ... ...
... ... @@ -56,9 +56,10 @@ print "UnitVelocity_in_cgs: ", const_unit_velocity_in_cgs # parameters of potential surface_density = 10. surface_density = 100. # surface density of all mass, which generates the gravitational potential scale_height = 100. gamma = 5./3. fgas = 0.1 # gas fraction # derived units const_unit_time_in_cgs = (const_unit_length_in_cgs / const_unit_velocity_in_cgs) ... ... @@ -131,7 +132,7 @@ h = glass_h[0:numGas] numGas = numpy.shape(pos)[0] # compute furthe properties of ICs column_density = surface_density * numpy.tanh(boxSize/2./scale_height) column_density = fgas * surface_density * numpy.tanh(boxSize/2./scale_height) enclosed_mass = column_density * boxSize * boxSize pmass = enclosed_mass / numGas meanrho = enclosed_mass / boxSize**3 ... ...
... ... @@ -97,7 +97,8 @@ //#define EXTERNAL_POTENTIAL_DISC_PATCH /* Source terms */ #define SN_FEEDBACK #define SOURCETERMS_NONE //#define SOURCETERMS_SN_FEEDBACK /* Cooling properties */ #define COOLING_NONE ... ...
... ... @@ -44,7 +44,7 @@ #include "potential.h" #include "runner.h" #include "scheduler.h" #include "sourceterms.h" #include "sourceterms_struct.h" #include "space.h" #include "task.h" #include "units.h" ... ...
... ... @@ -120,19 +120,11 @@ const char runner_flip[27] = {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, * @param timer 1 if the time is to be recorded. */ void runner_do_sourceterms(struct runner *r, struct cell *c, int timer) { struct part *restrict parts = c->parts; struct xpart *restrict xparts = c->xparts; const int count = c->count; const double cell_min[3] = {c->loc[0], c->loc[1], c->loc[2]}; const double cell_width[3] = {c->width[0], c->width[1], c->width[2]}; const int ti_current = r->e->ti_current; struct sourceterms *sourceterms = r->e->sourceterms; const double location[3] = {sourceterms->supernova.x, sourceterms->supernova.y, sourceterms->supernova.z}; const int dimen = 3; const double timeBase = r->e->timeBase; TIMER_TIC; ... ... @@ -143,78 +135,13 @@ void runner_do_sourceterms(struct runner *r, struct cell *c, int timer) { return; } /* is supernova still active? */ if (sourceterms->supernova.status == supernova_is_not_done) { /* does cell contain explosion? */ if (count > 0) { const int incell = is_in_cell(cell_min, cell_width, location, dimen); if (incell == 1) { /* inject SN energy into particle with highest id, if it is active */ int imax = 0; struct part *restrict p_sn = NULL; struct xpart *restrict xp_sn = NULL; for (int i = 0; i < count; i++) { /* Get a direct pointer on the part. */ struct part *restrict p = &parts[i]; if (p->id > imax) { imax = p->id; p_sn = p; xp_sn = &xparts[i]; } } if (count > 0) { /* Is this part within the time step? */ if (p_sn->ti_begin == ti_current) { /* Does this time step straddle the feedback injection time? */ const float t_begin = p_sn->ti_begin * timeBase; const float t_end = p_sn->ti_end * timeBase; if (t_begin <= sourceterms->supernova.time && t_end > sourceterms->supernova.time) { /* store old time step */ const int dti_old = p_sn->ti_end - p_sn->ti_begin; /* add supernova feedback */ do_supernova_feedback(sourceterms, p_sn); /* label supernova as done */ sourceterms->supernova.status = supernova_is_done; message(" applied super nova, time = %d, location= %e %e %e", ti_current, p_sn->x[0], p_sn->x[1], p_sn->x[2]); message(" applied super nova, velocity = %e %e %e", p_sn->v[0], p_sn->v[1], p_sn->v[2]); /* update timestep if new time step shorter than old time step */ const int dti = get_part_timestep(p_sn, xp_sn, r->e); if (dti < dti_old) { p_sn->ti_end = p_sn->ti_begin + dti; message(" changed timestep from %d to %d", dti_old, dti); /* apply simple time-step limiter on all particles in same cell: */ int i_limit = 0; for (int i = 0; i < count; i++) { struct part *restrict p = &parts[i]; const int dti_old = p->ti_end - p->ti_begin; if (dti_old > 2 * dti) { i_limit++; const int dti_new = 2 * dti; p->ti_end = p->ti_begin + dti_new; message(" old step = %d new step = %d", dti_old, dti_new); } else message(" old step = %d", dti_old); } message(" count= %d limited timestep of %d particles ", count, i_limit); } // error(" check! "); } } } /* do sourceterms in this cell? */ const int incell = sourceterms_test_cell(cell_min, cell_width, sourceterms, dimen); if (incell == 1) { sourceterms_apply(r, sourceterms, c); } } ... ...
... ... @@ -22,6 +22,7 @@ /* Local includes. */ #include "const.h" #include "hydro.h" #include "parser.h" #include "units.h" ... ... @@ -29,7 +30,7 @@ #include "sourceterms.h" /** * @brief Initialises the source terms * @brief Initialises the sourceterms * * @param parameter_file The parsed parameter file * @param us The current internal system of units ... ... @@ -37,30 +38,23 @@ */ void sourceterms_init(const struct swift_params* parameter_file, struct UnitSystem* us, struct sourceterms* source) { #ifdef SN_FEEDBACK source->supernova.time = parser_get_param_double(parameter_file, "SN:time"); source->supernova.energy = parser_get_param_double(parameter_file, "SN:energy"); source->supernova.x = parser_get_param_double(parameter_file, "SN:x"); source->supernova.y = parser_get_param_double(parameter_file, "SN:y"); source->supernova.z = parser_get_param_double(parameter_file, "SN:z"); source->supernova.status = supernova_is_not_done; #endif /* SN_FEEDBCK */ #ifdef SOURCETERMS_SN_FEEDBACK supernova_init(parameter_file, us, source); #endif /* SOURCETERMS_SN_FEEDBACK */ }; /** * @brief Prints the properties of the external potential to stdout. * * @brief Prints the properties of the source terms to stdout * @param source the structure that has all the source term properties */ void sourceterms_print(struct sourceterms* source) { #ifdef SN_FEEDBACK message( " Single SNe of energy= %e will explode at time= %e at location " "(%e,%e,%e)", source->supernova.energy, source->supernova.time, source->supernova.x, source->supernova.y, source->supernova.z); #endif /* SN_FEEDBACK */ #ifdef SOURCETERMS_NONE error(" no sourceterms defined yet you ran with -F"); #ifdef SOURCETERMS_SN_FEEDBACK #error can't have sourceterms when defined SOURCETERMS_NONE #endif #endif #ifdef SOURCETERMS_SN_FEEDBACK supernova_print(source); #endif /* SOURCETERMS_SN_FEEDBACK */ };
... ... @@ -18,22 +18,57 @@ ******************************************************************************/ #ifndef SWIFT_SOURCETERMS_H #define SWIFT_SOURCETERMS_H /** * @file src/sourceterms.h * @brief Branches between the different sourceterms functions. */ #include "./const.h" #ifdef SN_FEEDBACK #include "runner.h" #ifdef SOURCETERMS_SN_FEEDBACK #include "sourceterms/sn_feedback/sn_feedback_struct.h" #endif /* So far only one model here */ struct sourceterms { #ifdef SN_FEEDBACK #ifdef SOURCETERMS_SN_FEEDBACK struct supernova_struct supernova; #endif }; #ifdef SOURCETERMS_SN_FEEDBACK #include "sourceterms/sn_feedback/sn_feedback.h" #endif void sourceterms_init(const struct swift_params* parameter_file, struct UnitSystem* us, struct sourceterms* source); void sourceterms_print(struct sourceterms* source); #ifdef SN_FEEDBACK #include "sourceterms/sn_feedback/sn_feedback.h" /** * @file src/sourceterm.h * @brief Routines related to source terms * @param cell_min: corner of cell to test * @param cell_width: width of cell to test * @param sourceterms: properties of source terms to test * @param dimen: dimensionality of the problem * * This routine tests whether a source term should be applied to this cell * return: 1 if yes, return: 0 if no */ __attribute__((always_inline)) INLINE static int sourceterms_test_cell( const double cell_min[], const double cell_width[], struct sourceterms* sourceterms, const int dimen) { #ifdef SOURCETERMS_SN_FEEDBACK return supernova_feedback_test_cell(cell_min, cell_width, sourceterms, dimen); #endif return 0; }; __attribute__((always_inline)) INLINE static void sourceterms_apply( struct runner* r, struct sourceterms* sourceterms, struct cell* c) { #ifdef SOURCETERMS_SN_FEEDBACK supernova_feedback_apply(r, sourceterms, c); #endif }; #endif /* SWIFT_SOURCETERMS_H */
/******************************************************************************* * This file is part of SWIFT. * Coypright (c) 2015 Matthieu Schaller (matthieu.schaller@durham.ac.uk) * 2016 Tom Theuns (tom.theuns@durham.ac.uk) * * This program is free software: you can redistribute it and/or modify * it under the terms of the GNU Lesser General Public License as published * by the Free Software Foundation, either version 3 of the License, or * (at your option) any later version. * * This program is distributed in the hope that it will be useful, * but WITHOUT ANY WARRANTY; without even the implied warranty of * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the * GNU General Public License for more details. * * You should have received a copy of the GNU Lesser General Public License * along with this program. If not, see . * ******************************************************************************/ #include #include "feedback.h" /** * @brief Computes the feedback time-step of a given particle due to * a source term * * This function only branches towards the potential chosen by the user. * * @param feedback The properties of the source terms. * @param phys_const The physical constants in internal units. * @param g Pointer to the particle data. */ __attribute__((always_inline)) INLINE static float sourceterms_compute_timestep( const struct sourceterms* feedback, const struct phys_const* const phys_const, const struct part* const p) { float dt = FLT_MAX; #ifdef SN_FEEDBACK dt = fmin(dt, sn_feedback_timestep(feedback, phys_const, p)); #endif } __attribute__((always_inline)) INLINE static void feedback( const struct sourceterms* feedback, const struct phys_const* const phys_const, struct part* p) #ifdef SN_FEEDBACK sn_feedback(feedback, phys_const, p); #endif }
... ... @@ -19,40 +19,174 @@ #ifndef SWIFT_SN_FEEDBACK_H #define SWIFT_SN_FEEDBACK_H #include /* Config parameters. */ #include "../config.h" #include "engine.h" #include "equation_of_state.h" #include "hydro.h" /* determine whether location is in cell */ __attribute__((always_inline)) INLINE static int is_in_cell( const double cell_min[], const double cell_width[], const double location[], const int dimen) { #include "runner.h" #include "timestep.h" /** * @file src/sourceterms/sn_feedback.h * * @brief Routines related to sourceterms (supernova feedback): determine if * feedback occurs in this cell * * @param cell_min: corner of cell to test * @param cell_width: width of cell to test * @param sourceterms: properties of source terms to test * @param dimen: dimensionality of the problem * * This routine tests whether a source term should be applied to this cell * return: 1 if yes, return: 0 if no */ __attribute__((always_inline)) INLINE static int supernova_feedback_test_cell( const double cell_min[], const double cell_width[], struct sourceterms* sourceterms, const int dimen) { if (sourceterms->supernova.status == supernova_is_done) return 0; const double location[3] = {sourceterms->supernova.x, sourceterms->supernova.y, sourceterms->supernova.z}; for (int i = 0; i < dimen; i++) { if (cell_min[i] > location[i]) return 0; if ((cell_min[i] + cell_width[i]) <= location[i]) return 0; } }; return 1; }; /** * @file src/sourceterms/sn_feedback.h * @brief Routines related to source terms (supernova feedback) * * @brief Routines related to source terms (supernova feedback): perform * feedback in this cell * @param r: the runner * @param sourceterms the structure describing the source terms properties * @param p the particle to apply feedback to * @param c the cell to apply feedback to * * This routine heats an individual particle (p), increasing its thermal energy * per unit mass * by supernova energy / particle mass. */ __attribute__((always_inline)) INLINE static void do_supernova_feedback( const struct sourceterms* sourceterms, struct part* p) { const float u_old = hydro_get_internal_energy(p, 0); message(" u_old= %e entropy= %e", u_old, p->entropy); const float u_new = u_old + sourceterms->supernova.energy / hydro_get_mass(p); hydro_set_internal_energy(p, u_new); const float u_set = hydro_get_internal_energy(p, 0.0); message(" unew = %e %e s= %e", u_new, u_set, p->entropy); message( " injected SN energy in particle = %lld, increased energy from %e to %e, " "check= %e", p->id, u_old, u_new, u_set); __attribute__((always_inline)) INLINE static void supernova_feedback_apply( struct runner* restrict r, struct sourceterms* restrict sourceterms, struct cell* restrict c) { const int count = c->count; struct part* restrict parts = c->parts; struct xpart* restrict xparts = c->xparts; const double timeBase = r->e->timeBase; const int ti_current = r->e->ti_current; /* inject SN energy into the particle with highest id in this cell if it is * active */ int imax = 0; struct part* restrict p_sn = NULL; struct xpart* restrict xp_sn = NULL; for (int i = 0; i < count; i++) { /* Get a direct pointer on the part. */ struct part* restrict p = &parts[i]; if (p->id > imax) { imax = p->id; p_sn = p; xp_sn = &xparts[i]; } } /* Is this part within the time step? */ if (p_sn->ti_begin == ti_current) { /* Does this time step straddle the feedback injection time? */ const float t_begin = p_sn->ti_begin * timeBase; const float t_end = p_sn->ti_end * timeBase; if (t_begin <= sourceterms->supernova.time && t_end > sourceterms->supernova.time) { /* store old time step */ const int dti_old = p_sn->ti_end - p_sn->ti_begin; /* add supernova feedback */ const float u_old = hydro_get_internal_energy(p_sn, 0); message(" u_old= %e entropy= %e", u_old, p_sn->entropy); const float u_new = u_old + sourceterms->supernova.energy / hydro_get_mass(p_sn); hydro_set_internal_energy(p_sn, u_new); const float u_set = hydro_get_internal_energy(p_sn, 0.0); const float ent_set = hydro_get_entropy(p_sn, 0.0); message(" unew = %e %e s= %e", u_new, u_set, ent_set); message( " injected SN energy in particle = %lld, increased energy from %e to " "%e, " "check= %e", p_sn->id, u_old, u_new, u_set); /* label supernova as done */ sourceterms->supernova.status = supernova_is_done; message(" applied super nova, time = %d, location= %e %e %e", ti_current, p_sn->x[0], p_sn->x[1], p_sn->x[2]); message(" applied super nova, velocity = %e %e %e", p_sn->v[0], p_sn->v[1], p_sn->v[2]); /* update timestep if new time step shorter than old time step */ const int dti = get_part_timestep(p_sn, xp_sn, r->e); if (dti < dti_old) { p_sn->ti_end = p_sn->ti_begin + dti; message(" changed timestep from %d to %d", dti_old, dti); /* apply simple time-step limiter on all particles in same cell: */ int i_limit = 0; for (int i = 0; i < count; i++) { struct part* restrict p = &parts[i]; const int dti_old = p->ti_end - p->ti_begin; if (dti_old > 2 * dti) { i_limit++; const int dti_new = 2 * dti; p->ti_end = p->ti_begin + dti_new; message(" old step = %d new step = %d", dti_old, dti_new); } else message(" old step = %d", dti_old); } message(" count= %d limited timestep of %d particles ", count, i_limit); } /* end of limiter */ } } }; /** * @file src/sourceterms/sn_feedback.h * * @brief Routine to initialise supernova feedback * @param parameterfile: the parse parmeter file * @param us: the unit system in use * @param sourceterms the structure describing the source terms properties * * This routine heats an individual particle (p), increasing its thermal energy * per unit mass * by supernova energy / particle mass. */ __attribute__((always_inline)) INLINE static void supernova_init( const struct swift_params* parameter_file, struct UnitSystem* us, struct sourceterms* source) { source->supernova.time = parser_get_param_double(parameter_file, "SN:time"); source->supernova.energy = parser_get_param_double(parameter_file, "SN:energy"); source->supernova.x = parser_get_param_double(parameter_file, "SN:x"); source->supernova.y = parser_get_param_double(parameter_file, "SN:y"); source->supernova.z = parser_get_param_double(parameter_file, "SN:z"); source->supernova.status = supernova_is_not_done; } __attribute__((always_inline)) INLINE static void supernova_print( struct sourceterms* source) { message( " Single SNe of energy= %e will explode at time= %e at location " "(%e,%e,%e)", source->supernova.energy, source->supernova.time, source->supernova.x, source->supernova.y, source->supernova.z); } #endif /* SWIFT_SN_FEEDBACK_H */
... ... @@ -24,8 +24,8 @@ /* Local headers. */ #include "const.h" #include "cooling.h" #include "debug.h" /** * @brief Compute a valid integer time-step form a given time-step * ... ...
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Finish editing this message first! | 5,200 | 18,182 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2023-06 | latest | en | 0.43988 |
https://mattymatica.com/2020/10/09/the-law-of-universal-gravitation/ | 1,679,900,222,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948609.41/warc/CC-MAIN-20230327060940-20230327090940-00077.warc.gz | 446,829,134 | 55,138 | # The Law of Universal Gravitation
O Timothy! Guard what was committed to your trust, avoiding the profane and idle babblings and contradictions of what is falsely called knowledge—by professing it some have strayed concerning the faith. Grace be with you. Amen.
(1 Timothy 6:20-21) NKJV
Physical laws describe broad classes of observations. They’re a summary of what we know. They don’t define the universe, they’re defined by the universe.
If the universe was created, then the laws were defined by a creator. This is where the popular science freight train (SciPop) seems to have jumped the tracks. It got cause and effect wrong in its application of physical laws. Laws are an opportunistic rationalization of circumstantial evidence. Even though they may be shown to be consistent with all observations, they don’t reveal the underlying causes for the phenomena that they describe.
We all have the same evidence. Our choice of paradigm determines what we think it’s evidence of.
Matty’s Razor
Sir Isaac Newton wrote the law of universal gravitation. Here are the first two lines of what Wikipedia has to say about it:
Newton’s law of universal gravitation states that every particle attracts every other particle in the universe with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. This is a general physical law derived from empirical observations by what Isaac Newton called inductive reasoning.
– Gravity, definition Wikipedia
As the definition states, this is derived from empirical observation. There are many experiments which have been carried out which confirm the basic truth of this statement. The problem is that whatever experiments we do or measurements we make, they’re circumstantial evidence because they don’t tell us the cause of gravity. Newton is famously quoted as saying:
No great discovery was ever made without a bold guess.
– Sir Isaac Newton
Faith is believing in something that you can’t see, because of evidence.
– Faith, definition
The irony is that Newton didn’t make a great discovery, he made a bold guess and called it a discovery. In Newton’s case the bold guess was that mass is the cause of gravity. It must have seemed like a natural assumption to make. If the gravitational attraction of a body is proportional to its mass, then why wouldn’t mass be the cause of gravity? As straightforward as this may sound it’s wrong. It’s not a discovery, it’s the starting premise for a line of inductive reasoning. However it set the stage for the development of Newtonian Physics which is based on the idea that mass is the cause of gravity.
Inductive reasoning allowed Newton to design experiments which supported his premise. The results of the experiments are always going to support the premise, because the premise is based on observations of the phenomenon. It’s an example or circular reasoning.
• The hypothesis is used as the premise of an experiment,
• The experimental results confirm the hypothesis,
• The observation now has an explanation which is consistent with the premise used to design the experiment,
• The hypothesis is confirmed.
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# If 4 people can stain a deck in 2 days, how many people would be required to stain the deck in only 1 day?
A 8
Explaination
If it requires 4 people for 2 days, then it would require 8 people for 1 day.
B 2
C 4
D 12 | 86 | 266 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2021-21 | latest | en | 0.963206 |
https://www.sciencesfacts.com/solution/the-actual-income-for-this-month-has-been-reduced-200-1 | 1,701,412,341,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100276.12/warc/CC-MAIN-20231201053039-20231201083039-00417.warc.gz | 1,099,549,530 | 8,095 | # The actual income for this month has been reduced 200
2 months ago
## Solution 1
Guest #1422
2 months ago
No answer choices given but i think this is it.......This budget can be modified by increasing the amount spent on food and utilities and decreasing the amount spent on clothes and rent, thereby maintaining a positive actual net income.
## 📚 Related Questions
Question
F(x)=x 3 −6f, left parenthesis, x, right parenthesis, equals, x, start superscript, 3, end superscript, minus, 6 h(x)=\sqrt[\Large3]{2x-15}h(x)= 3 2x−15 h, left parenthesis, x, right parenthesis, equals, root, start index, 3, end index, square root of, 2, x, minus, 15, end square root Write f(h(x))f(h(x))f, left parenthesis, h, left parenthesis, x, right parenthesis, right parenthesis as an expression in terms of xxx. f(h(x))=f(h(x))=f, left parenthesis, h, left parenthesis, x, right parenthesis, right parenthesis, equals
Solution 1
For this case we have the following functions:
F (x) = x ^ 3-6
h (x) = 3 ^ root (2x-15)
Doing the composition of functions we have:
f (h (x)) = (3 ^ root (2x-15)) ^ 3-6
Rewriting we have:
f (h (x)) = 2x-15 - 6
f (h (x)) = 2x-21
The compound function f (h (x)) is given by:
f (h (x)) = 2x-21
Solution 2
2x - 21
......................
Question
How does tan 135 degrees compare to tan -135 degrees
Solution 1
Evaluating the tangent function for the two values we obtain the following:
tan 135=tan(135+180)=tan 315=-1
next:
tan -135=tan(180-135)=tan 45=1
thus comparing the two we see that tan 135 is negative while tan -135 is positive
Question
Solve the system of equations below x-2y=3 2y-3y=9
Solution 1
First you solve 2y-3y=9.
Now that you have y, then you substitute y into x-2y=3 to find X
Question
Instructions:Select the correct answer. Which function is the inverse of f(x)=50000(0.8)^x
Solution 1
To solve this problem you must follow the proccedure shown below:
1. You have the following function given in the problem:
f(x)=50000(0.8)^x
2. You can rewrite it as below:
y=
50000(0.8)^x
3. Then, you must apply interchange the variables:
x=50000(0.8)^y
4. Now, you have:
x/50000=(0.8)^y
5. When you apply ln you obtain:
ln(x/50000)=ln(0.8)^y
ln(x)-ln(50000)=yln(0.8)
6. Now, clear y:
y=[ln(x)-ln(50000)]/ln(0.8)
Question
"7 less than a number t" written as an algebraic expression is:
Solution 1
Answer: The expression for the given statement is
Step-by-step explanation:
We are given a statement:
7 less than a number t
Let the number be considered as 't'
'Less' in the above statement means subtraction operation.
The numerical value written before the 'less' operation is written after in the expression.
Thus, the expression for the given statement is
Question
How do you do a probability tree? How do you start it? How do I get all my percentages to add up to 100? Please help guys, this is for a project. I need to know the steps on how to do one or I'll funk the class.
Solution 1
why you just give the question give the question && the answer !!!
Solution 2
why you just give the question give the question && the answer !!!
Question
The truth table represents statements p, q, and r. p q r q ∧ r A T T T T B T T F F C T F T F D T F F F E F T T T F F T F F G F F T F H F F F F If p is false, which row represents when p ∨ (q ∧ r) is true?
Solution 1
Solution:
q ∧ r : q and r
q r q∧r
T T T
T F F
F T F
F F F
p ∨ (q∧r)
p OR (q AND r)
p q∧r p∨(q∧r)
F T∧ T=T T
F F∧T = F F
F T ∧ F=F F
F F ∧ F =F F
If P is false , The row which represents when p ∨ (q ∧ r) is true is →→F T T T
Which is Option E.
Solution 2
option e
Step-by-step explanation:
Question
An apple pie with a circumference of 28.26 inches is cut into 6 equal-sized pieces. Two pieces of the pie are eaten. What is the area of the pie that is left? Select the correct answer. Use 3.14 for π.
Solution 1
C = πd
d = C /
π
d = 28.26 / 3.14
d = 9 in
but r = d/2 = 9/2 = 4.5 in
A =
π r^2
= 3.14 x (4.5^2)
= 3.14 x 20.25
= 63.585
divided into 6 equal
so
63.585 / 6 = 10.5975
but ate 2, so there are 4 pieces left
4 x 10.5975 = 42.39
the area of the pie that is left is 42.36 in^2
Solution 2
Hey user!!
we know that circumference of circle = πd
given the circumference of the apple pie = 28.26 in.
therefore πd = 28.26 in.
==> 3.14 × d = 28.26 in.
==> d = 28.26/3.14
==> d = 9 in.
the diameter of the apple pie is 9 in.
therefore radius of the apple pie = 9/2 = 4.5 in.
area of the apple pie = πr²
== 3.14 × 4.5 × 4.5
= 63.585 in²
it's divided into 6 equal parts and two pieces are eaten. so the area of leftover pie = 63.585/6 × 4
= 63.585/3 × 2
= 21.195 × 2
= 42.39 in²
cheers!!
Question
Find the volume of the composite solid. Round your answer to the nearest tenth. Item 11 Find the volume of the composite solid. Round your answer to the nearest tenth.
Solution 1
We need a picture of the shape.
Solution 2
We need a picture of the shape.
Step-by-step explanation:
Question
Jonas is conducting an experiment using a 10-sided die. He determines that the theoretical probability of rolling a 3 is mc015-1.jpg. He rolls the die 20 times. Four of those rolls result in a 3. Which adjustment can Jonas make to his experiment so the theoretical and experimental probabilities are likely to be closer?
Solution 1
The link doesn't work, but I imagine he would have figured that there is a 1/10 chance of a 3 being rolled, assuming there are ten different numbers on the die.
Since he rolled the die 20 times, and 4 of those rolls were 3's, the experimental probability of him rolling a 3 are 1/5.
What I think he should do is: try the experiment 3 times, then gather the data of the experiments, and make his prediction.
Solution 2
Step-by-step explanation: its DDDDDDDDDD
2647929
842281
748681
586256
406852
368373
348603
324927
199835
130075
112100
106146
77164
23213
22589
19607
17108
13966
10987
3389 | 1,910 | 6,027 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2023-50 | latest | en | 0.736274 |
http://mygeekmonkey.com/2016/09/tcs-latest-question-paper-aptitude.html | 1,550,560,595,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247489425.55/warc/CC-MAIN-20190219061432-20190219083432-00198.warc.gz | 187,511,775 | 15,020 | # TCS Latest Question Paper Aptitude
### 1. The perimeter of a equilateral triangle and regular hexagon are equal. Find out the ratio of their areas?
a. 3:2
b. 2:3
c. 1:6
d. 6:1
Correct Option: b
Explanation:
Let the side of the equilateral triangle = aa units and side of the regular hexagon is bb units.
Given that, 3a=6b3a=6b ⇒ab=21⇒ab=21
Now ratio of the areas of equilateral triangle and hexagon = 3‾√4a2:33‾√2b234a2:332b2
⇒3‾√4(2)2:33‾√2(1)2⇒34(2)2:332(1)2
⇒2:3⇒2:3
### 2. What is the remainder of (32^31^301) when it is divided by 9?
a. 3
b. 5
c. 2
d. 1
Correct option: b
Explanation:
See solved example 6 here
3231301932313019 = 53130195313019
Euler totient theorem says that [aϕ(n)n]Rem=1[aϕ(n)n]Rem=1
ϕ(n)=n(1−1a)(1−1b)…ϕ(n)=n(1−1a)(1−1b)… here n=ap.bq…n=ap.bq…
Now ϕ(9)=9(1−13)=6ϕ(9)=9(1−13)=6
Therefore, 5656 when divided by 9 remainder 1.
Now 313016=1301=1313016=1301=1
So 3130131301 can be written as 6k + 1
⇒531301=(56)K.51⇒531301=(56)K.51
5313019=(56)K.519=1K.59=55313019=(56)K.519=1K.59=5
### 3. Which of the following numbers must be added to 5678 to give a reminder 35 when divided by 460?
a. 980
b. 797
c. 955
d. 618
Correct option: b
Explanation:
Let xx be the number to be added to 5678.
When you divide 5678 + xx by 460 the remainder = 35.
Therefore, 5678 + xx = 460k + 35 here kk is some quotient.
⇒⇒ 5643 + xx should exactly divisible by 460.
Now from the given options x = 797.
### 4. A girl entered a store and bought x flowers for y dollars (x and y are integers). When she was about to leave, the clerk said, “If you buy 10 more flowers I will give you all for \$\$2, and you will save 80 cents a dozen”. The values of x and y are:
a. (15,1)
b. (10,1)
c. (5,1)
d. Cannot be determined from the given information.
Correct option: c
Explanation:
Given she bought xx flowers for yy dollars.
So 1 flower cost = yxyx
12 flowers or 1 dozen cost = 12yx12yx
Again, xx+10 cost = 2 dollars
1 flower cost = 210+x210+x
12 flowers or 1 dozen cost = 2×1210+x=2410+x2×1210+x=2410+x
Given that this new dozen cost is 80 cents or 4/5 dollar less than original cost.
⇒12yx−2410+x=45⇒12yx−2410+x=45
From the given options, c satisfies this.
### 5. If a number is divided by 357 the remainder is 5, what will be the remainder if the number is divided by 17?
a. 9
b. 3
c. 5
d. 7
Correct option: c
Explanation:
Let ′N′′N′ be the given number.
N=357k+5N=357k+5 = 17×21k+517×21k+5
If this number is divided by 17 remainder is 5 as 357k is exactly divided by 17.
### 6. In how many possible ways can write 3240 as a product of 3 positive integers a,b and c.
a. 450
b. 420
c. 350
d. 320
Correct option:
Explanation:
3450=23×34×51=a×b×c3450=23×34×51=a×b×c
We have to distribute three 2’s to a, b, c in 3+3−1C3−1=5C2=103+3−1C3−1=5C2=10 ways
We have to distribute four 3’s to a, b, c in 3+4−1C3−1=6C2=153+4−1C3−1=6C2=15 ways
We have to distribute one 5 to a, b, c in 3 ways.
Total ways = 10×15×3=45010×15×3=450 ways.
### 7. On door A – It leads to freedomOn door B – It leads to Ghost houseOn door C – door B leads to Ghost houseThe statement written on one of the doors is wrong.Identify which door leads to freedom.
a. A
b. B
c. C
d. None
Correct option: c
Explanation:
Case 1: A, B are true. In this case, Statement C also correct. So contradiction.
Case
2: B, C are true. In this case, B leads to ghost house and C confirms
it. Now A is wrong. So door A does not lead to freedom. So Door C leads
to freedom.
### 8. In the given figure, If the sum of the values along each side is equal. Find the possible values a, b, c, d, e, and f.
a. 9, 7, 20, 16, 6, 38
b. 4, 9, 10, 13, 16, 38
c. 4, 7, 20, 13, 6, 38
d. 4, 7, 20, 16, 6, 33
Correct option: c
Explanation:
From the above table, 42 + a + b = 47 + e. Therefore, a + b = 5 + e. Option a, b ruled out.
47 + e = 15 + f. Therefore, 32 + e = f. Option d ruled out.
4 men throw a die each simultaneously. Find the probability that at least 2 people get the same number
a. 5/18
b. 13/18
c. 1/36
d. 1/2
### 9. 70, 54, 45, 41……. What is the next number in the given series?
a. 35
b. 36
c. 38
d. 40
Correct option: d
Explanation:
Consecutive squares are subtracted from the numbers.
70 – 54 = 16
54 – 45 = 9
45 – 41 = 4
So next we have to subtract 1. So answer = 41 – 1 = 40
### 10. How many positive integers less than 500 can be formed using the numbers 1,2,3,and 5 for digits, each digit being used only once.
a. 52
b. 68
c. 66
d. 34
Correct option:
Explanation:
Single digit number = 4
Double digit number = 4××3 = 12
Three digit numbers = 3××3××2= 18 (∵∵ If Hundred’s place is 5, then the number is greater than 500)
Total = 34.
Star mark question:
### 1. In particular language if A=0, B=1, C=2,…….. .. , Y=24, Z=25 then what is the value of ONE+ONE (in the form of alphabets only)
a. BDAI
b. ABDI
c. DABI
d. CIDA
Explanation:
This
problem is based on Base 26 rather than regular base 10 (decimal
system) that we normally use. In base 10 there are 10 digits 0 to 9
exist. In base 26 there are 26 digits 0 to 25 exist. To convert any
number into base 26, we have to divide the number with 26 and find the
remainder. (Study this Base system chapter).
Here, ONE + ONE =
E has value of 4. So E + E = 8 which is equal to I.
Now N + N = 13 + 13 = 26. But in base 26, there is no 26. So (26)10=(10)26(26)10=(10)26
So we put 0 and 1 carry over. But 0 in this system is A.
Now O + O + 1 = 14 + 14 + 1 = 29
Therefore, (29)10=(13)26(29)10=(13)26
But 1 = B and 3 = D in that system. So ONE + ONE = BDAI
### 2. Find the number of perfect squares in the given series 2013, 2020, 2027,……………., 2300 (Hint 44^2=1936)
a. 1
b. 2
c. 3
d. Can’t be determined
Explanation:
The
given series is an AP with common difference of 7. So the terms in the
above series are in the form of 2013 + 7k. We have to find the perfect
squares in this format in the given series.
Given that 44^2 = 1936.
Shortcut: To find the next perfect square, add 45th odd number to 44^2.
So 45^2 = 1936 + (2 x 45 -1) = 2025
46^2 = 2025 + (2 x 46 – 1) = 2116
47^2 = 2116 + (2 x 47 – 1) = 2209
Now subtract 2013 from the above numbers and divide by 7. Only 2209 is in the format of 2013 + 7k. One number satisfies.
### 3. What is in the 200th position of 1234 12344 123444 1234444….?
Explanation:
The given series is 1234, 12344, 123444, 1234444, …..
So the number of digits in each term are 4, 5, 6, … or (3 + 1), (3 + 2), (3 + 3), …..upto n terms = 3n+n(n+1)23n+n(n+1)2
So 3n+n(n+1)2≤2003n+n(n+1)2≤200
For
n = 16, We get 184 in the left hand side. So after 16 terms the number
of digits equal to 184. And 16 them contains 16 + 3 = 19 digits.
Now 17 term contains 20 digits and 123444……417times123444……4⏟17times. So last digit is 4 and last two digits are 44.
### 4. 2345 23455 234555 234555……….. what was last 2 numbers at 200th digit?
Explanation:
Proceed as above. The last two digits in the 200th place is 55.
### 5. There are equal number of boys and girls in a class. If 12 girls entered out, twice the boys as girls remain. What was the total number of students in a class?
Explanation:
Let the boys = b and girls = g
Given bg−12=21bg−12=21
Substitute b = g in the above equation. g = 24. So total students = 24 + 24 = 48
### 6. a bb ccc dddd eeeee ………What is the 120th letter?
Explanation:
Number of letters in each term are in AP. 1, 2, 3, …
So n(n+1)2≤120n(n+1)2≤120
For n = 15, we get LHS = 120. So 15th letter in the alphabet is O. So 15th term contains 15 O’s.
### 7. There are 120 male and 100 female in a society. Out of 25% male and 20% female are rural. 20% of male and 25% of female rural people passed in the exam. What % of rural students have passed the exam?
Explanation:
From the above data, Rural male = 25%(120) = 30, Rural female = 20%(100) = 20.
Passed students from rural: male = 20%(30) = 6, female = 25%(20) = 5
Required percentage = 1150×100=22%1150×100=22%
### 8. 1/7 th of the tank contains fuel. If 22 litres of fuel is poured into the tank the indicator rests at 1/5th mark. What is the quantity of the tank?
Explanation:
Let the tank capacity = vv liters.
Given, v7+22=v5v7+22=v5
v5−v7=22⇒v=385v5−v7=22⇒v=385
### 9. What is the probability of getting sum 3 or 4 when 2 dice are rolled
Explanation:
Required number of ways = (2, 1), (1, 2), (1, 3), (3, 1), (2, 2) = 5
Total ways = 62=3662=36
Probability = 536536 | 3,252 | 8,317 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2019-09 | latest | en | 0.676538 |
http://discourse.mc-stan.org/t/barriers-to-new-developers/163/4 | 1,521,766,959,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648113.87/warc/CC-MAIN-20180323004957-20180323024957-00416.warc.gz | 81,826,083 | 10,921 | # Barriers to New Developers
#1
This thread is for collecting various hindrances that people have encountered when trying to contribute to Stan for the first time in order to motivate better introductory materials. These can include, but are not limited to, undocumented or poorly-documented patterns, confusing style choices, and the like.
As an example, here are some comments that @Lu.Zhang made after working through the code for the first time.
I think there are three important skills for one to be a good coder for Stan:
The first is to know where I can find the function I need in Stan math library;
How it is coded in Stan; and How to use it in one’s own code. (Q1 & Q2)
The second is to understand functions specifically defined in Stan and learn how
to use them. (Q3 & Q4)
The third is to have a profound knowledge about programming in C++. (Q5 & Q6)
I made a small test out of the questions I had in the process of learning Stan code.
Hope this could help you to prepare for the interview.
1. What’s the difference and relationship between the following codes:
lib/eigen_3.2.9/Eigen/src/Cholesky/LLT.h
lib/eigen_3.2.9/Eigen/src/Cholesky/LDLT.h
lib/eigen_3.2.9/Eigen/src/Cholesky/LLT_MKL.h
stan/math/prim/mat/fun/cholesky_decompose.hpp
stan/math/prim/mat/fun/LDLT_factor.hpp
2.a) What is the most efficient way to calculate a multiplication of a lower-
triangular Cholesky factor with a vector? (No need to calculate determinant);
b)Try to list the least functions needed for checking error.
1. When to use “VectorViewMvt” and “VectorView”?
1. When to use “typedefs.hpp”? What is the usage of “partials_type.hpp”?
stan_math/fwd/mat/fun/typedefs.hpp
stan_math/fwd/scal/meta/partials_type.hpp
5.What is the function of “dynamic” in the following definition of Matrix?
using Eigen::Dynamic;
Matrix<double,Dynamic,2> coord2(1000, 2)
6.Try to input two 1000 by 4 matrics A and B, and a 4 by 4 matrix C from a text file,
then use functions in stan library or any other available library to calculate: A + B *C^-1.
#2
Were all the questions answered?
I think what we need is something like a high-level overview,
and maybe a case study that walks someone through all the
steps to contributing. Lu’s questions are diving into specific
details, which would make most sense in a case study.
• Bob
#3
Are those supposed to be screening questions for somebody who wants to develop with Stan? They’re confusing.
#4
I like the idea of having this thread. It would be fun at some point to teach a workshop on this to get more people into working with the core of Stan.
Some issues I’ve noticed:
Lack of time (on our parts) to properly address design issues with potential contributions. It’s hard over email/github to sort them out and new contributors often don’t understand what our priorities are in terms of efficiency and maintainability. Sebastian and I both had the issue where we wrote a bunch of code while in communication with Daniel/Bob (not pointing fingers, you guys were just the ones willing to correspond over design issues) but in the end a lot of code hard to be re-written anyway due to misunderstandings. It would be nice to flesh out some of our priorities in a doc for new developers. The things I can think of are:
1. most of the time is spent in auto-diff so maintainability/UI should often win over efficiency in non auto-diff code.
2. maintainability is critical so don’t try to future-proof by adding indirection
3)… I forget, I think I’ve gotten used to the project so I lost perspective on these…
There are also some basic tasks that could use examples, for example:
1. example of adding a function to stan-dev/math, I think we have this in some form
2. example of exposing a function to stan-dev/stan
3. example of adding a function to stan-dev/math, with analytic derivatives (In the next few weeks I want to add boost’s inverse gamma so I’ll try to track the process and produce a simple example).
4. pointers to tests that demonstrate how to call the stan-dev/stan API for algorithms, these don’t necessarily need to be separate examples since the tests Dan and I wrote on the refactor branch are pretty clear. It would help to have some of the objects available.
5. maybe a generated C++ file for a Stan model with commentary (possibly just in comments) to clear up the ‘model concept’ we’re using. This stuff didn’t really make sense to me until I tried to modify the code generator.
All these pieces make sense to me now but it used to look like a black art. I don’t think much of it really requires a deep understanding of C++, although writing a function with analytic gradients requires more than the rest.
Multivariate Function with Known Gradients - RStan
#5
Krzysztof:
Feel free to write those docs!
I think we already have have everything you’re asking
for in bits and pieces, but not in the form of a coherent
document.
I thought all the code we currently have in Stan would
work as an example. Or do you mean something like a step-by-step
example? It’s always hard to know when to stop. We might need
more than one of these docs for people at different levels of
understanding of Git, C++, matrix arithmetic, etc.
• Bob
#6
Also anyone can feel free to post something on the blog with this material. I think it will be of general interest, and not just for Stan people.
A
#7
Sure. This is meant to be an inclusive thread aggregating confusions people have had when they started and they may very well lead to numerous docs to guide new developers into the project.
#8
No, not screening questions. Just confusions/speed bumps new developers have had when they started looking at the code. The idea is to identify what should be communicated to new developers to make their transition as smooth as possible.
#9
What I meant was a case study of how to write a particular
function. Like a video of all the steps involved.
It’s always nice to see a concrete example in addition to
high-level pointers to Git doc and books on C++ template
programming.
• Bob
#10
That would be cool. I’d sign up for (attending) that.
#11
I’m feeling the freedom! Just the time hasn’t come up yet.
I’m thinking on the level of what it would take to get a committed undergrad with some coding experience and the required (minimum) math background to walk through tasks 1-4. For C++ concepts we can send people to basic books, for math we can ask that they know what a partial derivative is, but I’m not expecting people to grok why the calculation of gradients is set up the way it is by reading the code. Instead I’d like to point them to the right place to plug in the relevant partial derivative. You’re right we have a lot of this doc (and the Stan autodiff paper explains it really well) but I’d like to take the relevant pieces and put them into a worked example.
We basically did this on the Wiki for #1 and #2, I’d like to polish those and find a way to integrate them into workshop-style material (maybe a github repo with Markdown/C++ examples, etc…)
#12
sakrejda Developer
December 16
Bob_Carpenter:
Feel free to write those docs!
I’m feeling the freedom! Just the time hasn’t come up yet.
Bob_Carpenter:
I thought all the code we currently have in Stan would work as an example. Or do you mean something like a step-by-step example?
I’m thinking on the level of what it would take to get a committed undergrad with some coding experience and the required (minimum) math background to walk through tasks 1-4. For C++ concepts we can send people to basic books, for math we can ask that they know what a partial derivative is, but I’m not expecting people to grok why the calculation of gradients is set up the way it is by reading the code. Instead I’d like to point them to the right place to plug in the relevant partial derivative. You’re right we have a lot of this doc (and the Stan autodiff paper explains it really well) but I’d like to take the relevant pieces and put them into a worked example.
We basically did this on the Wiki for #1 and #2, I’d like to polish those and find a way to integrate them into workshop-style material (maybe a github repo with Markdown/C++ examples, etc…)
That sounds great. I think you’ll find the undergrads
will vary in background among all the relevant factors.
We’d
like to be able to enable people who haven’t used GitHub or
C++. It’s a daunting learning curve and an example would be super
great.
What I think makes sense is
• clone the repo and get the build working
• create an issue, create a branch, add a test that fails,
fix the bug, create a pull request
• add a new templated function to Stan math
• push a function through Stan (submodules) with doc (LaTeX)
• add a new function with analytic partial derivatives
• add a new distribution with all the template metaprogramming
There’s other contributions to make up at the language, interface
and algorithm level, but I think it makes sense to tackle the
above at the math and Stan lib level first.
I think we could do the same for dealing with the R and Python
interfaces. Or working up at a higher level, like adding a new
language feature.
• Bob
#13
Hey,
I’m trying to get started myself and have had a few issues so far. My background - “industrial programmer,” no prior professional C++ experience but have worked in 10 other languages.
Stuff I’ve struggled with:
1. Where should I be asking questions for this kind of stuff? Some orgs have an IRC room or something similar that serves as triage. Should be something you don’t need to get permission to post on. And some of the questions can be a bit embarrassing for a long lived forum post like this.
2. Getting my dev environment set up. Spent the morning so far trying to configure emacs to understand where the header files for stan-dev/math live. Generally - how should I check out the various repos? Can I just clone one that clones the others as submodules? Where do compiler or IDE settings live for each? Maybe someone casually posting their emacs config would be useful as a starter.
3. Build process - how do I know I’ve entered a known-good build state, so that I can then make changes on top of that?
Currently spending a lot of time on #2, and given #1 I’m not sure if this is the right place to also post asking for help with that :P
Thanks,
Sean
#14
This or the users list is the right place, happy to help out as much as possible… in the context of the large pile of questions that we’ve all asked I don’t think you could come up with anything all that embarrassing.
git clone the develop branch
Clone the one you are working on. If your really good with git submodules you might get away with cloning just CmdStan and then correctly handling submodules if you’re working with, e.g., stan-dev/math, but that’s making it more complicated than it needs to be.
There’s a make subdirectory for compiler-related stuff, any IDE-related stuff ends up in a separate repo (?)
Sorry, I use vim… :)
Develop is always in a good build state.
#15
Vim config also useful! I will use whichever one starts working more quickly - in emacs I use evil-mode anyway.
re: building - I meant literally “what are the steps to build each of these repos?” I’m trying to add tests to the math repo and there are no make targets that seem to make sense for building, so I tried running runTests.py which I suspect will build as a side effect, but that command has been running for a long time and hasn’t finished yet so I’m still not sure if it was the correct entry point.
w.r.t. compiler settings - I think I might understand now how the vendored libraries are included, but some brief description of why the project is laid out the way it is with respect to directory structure, which pieces are dependencies of which other pieces, how dependencies are included, that kind of thing would be very welcome. Even if a veteran C++ developer might understand some of these things by reading the source, we probably want to lower that barrier to entry.
#16
sakrejda Developer
December 16
seantalts:
• Where should I be asking questions for this kind of stuff? Some orgs have an IRC room or something similar that serves as triage. Should be something you don’t need to get permission to post on. And some of the questions can be a bit embarrassing for a long lived forum post like this.
No IRC because nobody wants to deal with a constant flow of
interruptions. (Well maybe somebody does, but I don’t.)
This or the users list is the right place, happy to help out as much as possible… in the context of the large pile of questions that we’ve all asked I don’t think you could come up with anything all that embarrassing.
seantalts:
Generally - how should I check out the various repos?
git clone the develop branch
The developer process wiki explains our whole process. It’s
on stan-dev/stan. This is the stuff we’re talking about
consolidating.
seantalts:
Can I just clone one that clones the others as submodules?
Clone the one you are working on. If your really good with git submodules you might get away with cloning just CmdStan and then correctly handling submodules if you’re working with, e.g., stan-dev/math, but that’s making it more complicated than it needs to be.
I start with cmdstan, then from cmdstan you can run:
make stan-update
which will pull out the stan submodule. Then cd into
stan and run
make math-update
which will pull in the math lib.
Then don’t change those and don’t commit new submodule links
directly—we handle those updates through automatically generated
pull requests from the continuous integration system when an
upstream module changes.
seantalts:
Where do compiler or IDE settings live for each?
There’s a make subdirectory for compiler-related stuff, any IDE-related stuff ends up in a separate repo (?)
We just have makefiles. There’s no IDE stuff?
There’s also no compiler-related stuff built into Stan. We
just use the system call or you can create a make.local to
override. We tend to use clang++ on the Mac from Xcode
with O=0 for optimization level 0 during builds and testing.
seantalts:
Maybe someone casually posting their emacs config would be useful as a starter.
Sorry, I use vim… :)
Mine’s out of date because I hate installing software.
Everything other than how to configure stan-mode is in
our Developer Tricks wiki.
seantalts:
• Build process - how do I know I’ve entered a known-good build state, so that I can then make changes on top of that?
Develop is always in a good build state.
We try. You can run the unit tests to make sure.
• Bob
#17
I don’t really have one… every once in a while Daniel tells me I need to remove whitespace at the end of lines of C++ and I have to break out sed but … I think you just need to wait for one of people who’s more into IDE’s to answer.
The math library is header-only so you never build it as a shared library.
That does build and run tests, and you can pass it specific test files to run, but it doesn’t build a library.
Good point, one of us should really do that…
#18
seantalts Developer
December 16
Vim config also useful! I will use whichever one starts working more quickly - in emacs I use evil-mode anyway.
re: building - I meant literally “what are the steps to build each of these repos?” I’m trying to add tests to the math repo and there are no make targets that seem to make sense for building, so I tried running runTests.py which I suspect will build as a side effect, but that command has been running for a long time and hasn’t finished yet so I’m still not sure if it was the correct entry point.
The math library is primarily header only. The only thing
that gets build in the sense of generating object code is the
ODE solver library.
You can do something like this:
make -j4 O=0 CC=clang++ test/unit
which will kick off all the unit tests using 4 cores using optimization
level 0 and the clang compiler.
w.r.t. compiler settings - I think I might understand now how the vendored libraries are included, but some brief description of why the project is laid out the way it is with respect to directory structure, which pieces are dependencies of which other pieces, how dependencies are included, that kind of thing would be very welcome.
All there on the wiki. Just not well organized. You
want to start from the top level Wiki page, which is a directory:
You’ll see the layout of the repos there. If you click through
the process, you’ll get more detailed layout of directories.
The math lib has its own wiki:
Somewhere, there’s a list of dependency order among scalar,
array and matrix, and primitive, reverse-mode, forward-mode
and mixed autodiff.
Even if a veteran C++ developer might understand some of these things by reading the source, we probably want to lower that barrier to entry.
Absolutely. Very high priority for the project, I think.
• Bob
#19
(defun java-mode-untabify ()
(save-excursion
(goto-char (point-min))
(if (search-forward “t” nil t)
(untabify (1- (point)) (point-max))))
nil)
’(lambda ()
(make-local-variable 'write-contents-hooks)
’(lambda ()
(make-local-variable 'write-contents-hooks)
’(lambda ()
(make-local-variable 'write-contents-hooks)
’(lambda ()
(make-local-variable 'write-contents-hooks)
(setq indent-tabs-mode nil)
(put 'downcase-region 'disabled nil)
(require 'stan-mode)
(setq x-select-enable-clipboard t)
(lambda () (add-to-list 'write-file-functions 'delete-trailing-whitespace)))
The last line gets rid of the trailing whitespace. There’s
also code to use the StanMode, but mine’s out of date, I’m sure.
There’s also the tab killing code in here.
• Bob
#20
sakrejda Developer
December 16
seantalts:
Vim config also useful! I will use whichever one starts working more quickly - in emacs I use evil-mode anyway.
I don’t really have one… every once in a while Daniel tells me I need to remove whitespace at the end of lines of C++ and I have to break out sed but … I think you just need to wait for one of people who’s more into IDE’s to answer.
Run:
make cpplint
which runs Google’s code style checker. We’re down to zero errors.
There’s also reference to that and some exceptions in the Stan repo
wiki.
• Bob | 4,165 | 18,197 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2018-13 | latest | en | 0.867179 |
https://whatisconvert.com/77-square-meters-in-square-yards | 1,600,493,233,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400190270.10/warc/CC-MAIN-20200919044311-20200919074311-00280.warc.gz | 705,540,685 | 7,325 | # What is 77 Square Meters in Square Yards?
## Convert 77 Square Meters to Square Yards
To calculate 77 Square Meters to the corresponding value in Square Yards, multiply the quantity in Square Meters by 1.1959900463011 (conversion factor). In this case we should multiply 77 Square Meters by 1.1959900463011 to get the equivalent result in Square Yards:
77 Square Meters x 1.1959900463011 = 92.091233565183 Square Yards
77 Square Meters is equivalent to 92.091233565183 Square Yards.
## How to convert from Square Meters to Square Yards
The conversion factor from Square Meters to Square Yards is 1.1959900463011. To find out how many Square Meters in Square Yards, multiply by the conversion factor or use the Area converter above. Seventy-seven Square Meters is equivalent to ninety-two point zero nine one Square Yards.
## Definition of Square Meter
The square metre (International spelling as used by the International Bureau of Weights and Measures) or square meter (American spelling) is the SI derived unit of area, with symbol m2 (33A1 in Unicode). It is defined as the area of a square whose sides measure exactly one metre.
## Definition of Square Yard
The square yard is an imperial unit of area, formerly used in most of the English-speaking world but now generally eplaced by the square metre, however it i still in widespread use in the US., Canada and the U.K. It isdefined as the area of a quare with sides of one yard (three feet, thirty-six inches, 0.9144 metres) in length. There is no universally agreed symbol but the following are used: square yards, square yard, square yds, square yd, sq yards, sq yard, sq yds, sq yd, sq.yd., yards/-2, yard/-2, yds/-2, yd/-2, yards^2, yard^2, yds^2, yd^2, yards², yard², yds², yd².
## Using the Square Meters to Square Yards converter you can get answers to questions like the following:
• How many Square Yards are in 77 Square Meters?
• 77 Square Meters is equal to how many Square Yards?
• How to convert 77 Square Meters to Square Yards?
• How many is 77 Square Meters in Square Yards?
• What is 77 Square Meters in Square Yards?
• How much is 77 Square Meters in Square Yards?
• How many yd2 are in 77 m2?
• 77 m2 is equal to how many yd2?
• How to convert 77 m2 to yd2?
• How many is 77 m2 in yd2?
• What is 77 m2 in yd2?
• How much is 77 m2 in yd2? | 641 | 2,327 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2020-40 | latest | en | 0.915498 |
https://dev.to/theabbie/longest-valid-parentheses-1o3k | 1,721,901,219,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763857355.79/warc/CC-MAIN-20240725084035-20240725114035-00281.warc.gz | 182,260,032 | 75,409 | ## DEV Community
Abhishek Chaudhary
Posted on
# Longest Valid Parentheses
Given a string containing just the characters `'('` and `')'`, find the length of the longest valid (well-formed) parentheses substring.
Example 1:
Input: s = "(()"
Output: 2
Explanation: The longest valid parentheses substring is "()".
Example 2:
Input: s = ")()())"
Output: 4
Explanation: The longest valid parentheses substring is "()()".
Example 3:
Input: s = ""
Output: 0
Constraints:
• `0 <= s.length <= 3 * 104`
• `s[i]` is `'('`, or `')'`.
SOLUTION:
``````class Solution:
def longestValidParentheses(self, s: str) -> int:
n = len(s)
stack = [-1]
mlen = 0
for i, c in enumerate(s):
if c == '(':
stack.append(i)
elif len(stack) > 0:
stack.pop()
if len(stack) > 0:
mlen = max(mlen, i - stack[-1])
else:
stack.append(i)
return mlen
`````` | 251 | 830 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-30 | latest | en | 0.540033 |
https://mphitchman.com/geometry/section5-3.html | 1,714,027,016,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712297290384.96/warc/CC-MAIN-20240425063334-20240425093334-00065.warc.gz | 357,511,165 | 11,820 | ## Section5.3Measurement in Hyperbolic Geometry
In this section we develop a notion of distance in the hyperbolic plane. If someone is standing at point $p$ and wants to get to point $q\text{,}$ he or she should be able to say how far it is to get there, whatever the route taken.
The distance formula is derived following the approach given in Section 30 of Boas' text [2]. We first list features our distance function ought to have, and use the notation that $d_H(p,q)$ represents the hyperbolic distance from $p$ to $q$ in the hyperbolic plane $\mathbb{D}\text{.}$
1. The distance between 2 distinct points should be positive.
2. The shortest path between 2 points should be on the hyperbolic line connecting them.
3. If $p,q\text{,}$ and $r$ are three points on a hyperbolic line with $q$ between the other two then $d_H(p,q) + d_H(q,r) = d_H(p,r)\text{.}$
4. Distance should be preserved by transformations in $\cal H\text{.}$ (A lunch pail shouldn't shrink if it is moved to another table.) In other words, the distance formula should satisfy
\begin{equation*} d_H(p,q) = d_H(T(p),T(q)) \end{equation*}
for any points $p$ and $q$ in $\mathbb{D}\text{,}$ and any transformation $T$ in ${\cal H}\text{.}$
5. In the limit for small distances, hyperbolic distance should be proportional to Euclidean distance.
Perhaps the least obvious of the features listed is the last one. One theme of this text is that locally, on small scales, non-Euclidean geometry behaves much like Euclidean geometry. A small segment in the hyperbolic plane is approximated to the first order by a Euclidean segment. Small hyperbolic triangles look like Euclidean triangles and hyperbolic angles correspond to Euclidean angles; the hyperbolic distance formula will fit with this theme.
To find the distance function, start with a point's distance from the origin. Given a point $z$ in $\mathbb{D}\text{,}$ rotate about 0 so that $z$ gets sent to the point $x = |z|$ on the positive real axis.
We may find a hyperbolic line $L$ about which $x$ gets reflected to the origin. Such a hyperbolic line is constructed in the proof of Theorem 5.1.3. Recall, the line $L$ is on the circle centered at $x^*$ (the point symmetric to $x$ with respect to to $\mathbb{S}^1_\infty$) that goes through the points at which $\mathbb{S}^1_\infty$ intersects the circle with diameter $0x^*\text{.}$ Let $x + h$ be a point near $x$ on the positive real axis, and suppose $x + h$ gets inverted to the point $w\text{,}$ as depicted in Figure 5.3.1. One can show (in Exercise 5.3.1) that
\begin{equation*} w = \frac{-h}{1 - x^2 - hx}\text{.} \end{equation*}
If distance is to be preserved by transformations in ${\cal H}\text{,}$
\begin{equation*} d_H(x,x+h) = d_H(0,w) \tag{1}\text{.} \end{equation*}
Also, $0, x\text{,}$ and $x+h$ are all on the same hyperbolic line (the real axis), so assuming $h > 0$
\begin{equation*} d_H(0,x)+d_H(x,x+h) = d_H(0,x+h) \tag{2}\text{.} \end{equation*}
For $x \in \mathbb{R}$ define the function $d(x) = d_H(0,x)$ which is the hyperbolic distance of $x$ to the origin. Then (2) and (1) may be combined to give
\begin{equation*} d(x+h)-d(x)=d(w)\text{.} \end{equation*}
Divide both sides by $h$ to get
\begin{equation*} \frac{d(x+h)-d(x)}{h}=\frac{d(w)}{h}\text{.} \end{equation*}
As $h \to 0$ we obtain
\begin{equation*} d^\prime(x) = \lim_{h \to 0} \frac{d(w)}{h}\text{.} \end{equation*}
We now interrupt this derivation with an important point. In the limit for small $w\text{,}$ the hyperbolic distance of $w$ from 0, $d(w)\text{,}$ is proportional to the Euclidean distance $|w - 0| = |w|\text{.}$ Since $w$ is the image of $x+h$ under the inversion and $x$ gets inverted to 0, it follows that $w \to 0$ as $h \to 0\text{.}$ So, we assume that
\begin{equation*} \lim_{h \to 0} \frac{d(w)}{|w|} = k \end{equation*}
for some constant $k\text{.}$ Following convention, we set the constant of proportionality to $k = 2\text{,}$ as this makes length and area formulas look very nice later on. Now, back to the derivation.
\begin{align*} d^\prime(x) \amp = \lim_{h \to 0} \frac{d(w)}{h}\\ \amp = \lim_{h \to 0} \frac{d(w)}{|w|}\frac{|w|}{h}\\ \amp = \lim_{h \to 0} \bigg[ 2 \cdot \frac{h}{(1-x^2-hx)h} \bigg]\\ \amp = \frac{2}{1-x^2}\text{.} \end{align*}
To get back to the distance function $d(x)$ we integrate:
\begin{align*} \int \frac{2}{1-x^2} ~dx \amp = \int \bigg( \frac{1}{1-x} + \frac{1}{1+x}\bigg)~dx \tag{partial~fractions}\\ \amp = -\ln(1-x) + \ln(1+x)\\ d(x) \amp = \ln\left(\frac{1+x}{1-x}\right)\text{,} \end{align*}
so we arrive at the following distance formula.
###### The hyperbolic distance from 0 to $z$.
The hyperbolic distance from 0 to a point $z$ in $\mathbb{D}$ is
\begin{equation*} d_H(0,z) = \ln\left(\frac{1+|z|}{1-|z|}\right)\text{.} \end{equation*}
Notice that if $z$ inches its way in $\mathbb{D}$ out toward the circle at infinity (i.e., $|z| \to 1$), the hyperbolic distance from $0$ to $z$ approaches $\infty\text{.}$ This is a good thing. Thinking of Euclid's postulates, this notion of distance satisfies one of our fundamental requirements: One can produce a hyperbolic segment to any finite length.
To arrive at a general distance formula $d_H(p,q)\text{,}$ observe something curious. The hyperbolic line through $0$ and $x$ has ideal points -1 and 1. Furthermore, the expression $(1+x)/(1-x)$ corresponds to the cross ratio of the points $0\text{,}$ $x\text{,}$ 1, and -1. In particular,
\begin{equation*} (0,x;1,-1) = \frac{0 - 1}{0 + 1}\cdot \frac{x + 1}{x - 1} = \frac{1 + x}{1 - x}\text{.} \end{equation*}
Thus,
\begin{equation*} d_H(0,x) = \ln((0,x;1,-1))\text{.} \end{equation*}
We can now derive a general distance formula, assuming the invariance of distance under transformations in ${\cal H}\text{.}$ There is a transformation $T$ in ${\cal H}$ that takes $p$ to the origin and $q$ to some spot on the positive real-axis, call this spot $x$ (see Figure 5.3.2). Thus,
\begin{align*} d_H(p,q) \amp = d_H(T(p),T(q)) \tag{invariance of distance}\\ \amp = d_H(0,x)\\ \amp = \ln((0,x;1,-1))\\ \amp = \ln((p,q;u,v))\tag{invariance of cross ratio}\text{,} \end{align*}
where $u$ and $v$ are the ideal points of the hyperbolic line through $p$ and $q\text{.}$ To be precise, $u$ is the ideal point you would head toward as you went from $p$ to $q\text{,}$ and $v$ is the ideal point you would head toward as you went from $q$ to $p\text{.}$
### SubsectionA working formula for $\mathbf{d_H(p,q)}$
One may compute the hyperbolic distance between $p$ and $q$ by first finding the ideal points $u$ and $v$ of the hyperbolic line through $p$ and $q$ and then using the formula $d_H(p,q)=\ln((p,q;u,v))\text{.}$ In practice, finding coordinates for these ideal points can be a difficult task, and it is often simpler to compute the distance between points by first moving one of them to the origin. (This simpler approach uses the fact that hyperbolic distance is preserved under transformations in $\cal H\text{.}$ This fact will be proved shortly.)
One transformation in ${\cal H}$ that sends $p$ to 0 has the form
\begin{equation*} T(z) = \frac{z-p}{1-\overline{p}z}\text{.} \end{equation*}
The map $T$ sends $q$ to some other point, $T(q)\text{,}$ in $\mathbb{D}\text{.}$ Assuming again that $T$ preserves distance, it follows that $d_H(p,q) = d_H(0,T(q))\text{,}$ and
\begin{equation*} d_H(p,q) = \ln\left(\frac{1+|T(q)|}{1-|T(q)|}\right)\text{.} \end{equation*}
Making the substitution $\displaystyle T(q) = \frac{q-p}{1-\overline{p}q}$ provides us with the following working formula for the hyperbolic distance between two points.
###### Example5.3.4.The distance between two points.
For instance, suppose $p = \frac{1}{2}i\text{,}$ $q = \frac{1}{2} + \frac{1}{2}i\text{,}$ $z = .95e^{i5\pi/6}$ and $w=-.95\text{.}$ Then $d_H(p,q) \approx 1.49$ units, while $d_H(z,w) \approx 4.64$ units.
### SubsectionThe arc-length differential
Now that we can compute the distance between two points in the hyperbolic plane, we turn our attention to measuring the length of any path that takes us from $p$ to $q\text{.}$
###### Definition5.3.5.
A smooth curve is a differentiable map from an interval of real numbers to the plane
\begin{equation*} \boldsymbol{r}: [a,b] \to \mathbb{C} \end{equation*}
such that $\boldsymbol{r}^\prime(t)$ exists for all $t$ and never equals $0\text{.}$
In the spirit of this text, we write $\boldsymbol{r}(t) = x(t) + i y(t)\text{,}$ in which case $\boldsymbol{r}^\prime(t) = x^\prime(t) + iy^\prime(t)\text{.}$
Recall that in calculus we first approximate the Euclidean length of a given smooth curve $\boldsymbol{r}(t) = x(t) + iy(t)$ by summing the contributions of small straight line segments having Euclidean length
\begin{align*} \Delta s \amp = |\boldsymbol{r}(t+\Delta t) - \boldsymbol{r}(t)|\\ \amp = \sqrt{[x(t + \Delta t) - x(t)]^2 + [y(t + \Delta t) - y(t)]^2}\\ \amp = \sqrt{\bigg[ \frac{x(t + \Delta t) - x(t)}{\Delta t}\bigg]^2 + \bigg[ \frac{y(t + \Delta t) - y(t)}{\Delta t}\bigg]^2} |\Delta t|\text{.} \end{align*}
As $\Delta t \to 0$ we obtain the Euclidean arc-length differential
\begin{equation*} ds = \sqrt{(dx/dt)^2 + (dy/dt)^2}~dt\text{,} \end{equation*}
which may be expressed as
\begin{equation*} ds = |\boldsymbol{r}^\prime(t)|~dt\text{.} \end{equation*}
For instance, we may compute the (Euclidean) circumference of a circle with radius $a$ as follows. Consider $\boldsymbol{r}: [0, 2\pi] \to \mathbb{C}$ by $\boldsymbol{r}(t) = a \cos(t) + i a \sin(t)\text{.}$ This map traces a circle of radius $a$ centered at the origin. To find the length of this curve, which we denote as ${\cal L}(\boldsymbol{r})\text{,}$ compute the integral
\begin{align*} {\cal L}(\boldsymbol{r}) \amp = \int_0^{2\pi} |\boldsymbol{r}^\prime(t)|~dt\\ \amp = \int_0^{2\pi} |-a\sin(t) + i a \cos(t)|~dt\\ \amp = \int_0^{2\pi}\sqrt{a^2 \sin^2(t) + a^2 \cos^2(t)}~dt\\ \amp = \int_0^{2\pi} a ~dt\\ \amp = 2\pi a\text{.} \end{align*}
In the hyperbolic plane, we may deduce the arc-length differential by a similar argument. Suppose $\boldsymbol{r}$ is a smooth curve in $\mathbb{D}$ given by $\boldsymbol{r}(t) = x(t) + iy(t)\text{,}$ for $a \leq t \leq b\text{.}$ One may approximate the length of a tiny portion of the curve, say from $\boldsymbol{r}(t)$ to $\boldsymbol{r}(t + \Delta t)\text{,}$ by the hyperbolic distance between these two points, $d_H(\boldsymbol{r}(t), \boldsymbol{r}(t + \Delta t))\text{.}$ To compute this distance, we first send the point $\boldsymbol{r}(t)$ to 0 by the transformation
\begin{equation*} T(z) = \frac{z-\boldsymbol{r}(t)}{1-\overline{\boldsymbol{r}(t)}z}\text{,} \end{equation*}
so that
\begin{equation*} d_H(\boldsymbol{r}(t),\boldsymbol{r}(t+\Delta t)) = \ln[1 + |T(\boldsymbol{r}(t+\Delta t))|]-\ln[1 - |T(\boldsymbol{r}(t+\Delta t))|]\text{.} \end{equation*}
To arrive at an arc-length differential, we want to let $\Delta t$ approach 0. As this happens, $T(\boldsymbol{r}(t+\Delta t))$ approaches $T(\boldsymbol{r}(t))\text{,}$ which is 0. From calculus we also know that $\ln(1+x) \approx x$ for $x$ very close to 0. Thus, for small $\Delta t\text{,}$ we have
\begin{align*} d_H(\boldsymbol{r}(t),\boldsymbol{r}(t+\Delta t)) \amp \approx |T(\boldsymbol{r}(t+\Delta t))| + |T(\boldsymbol{r}(t+\Delta t))|\\ \amp =2\cdot \left| \frac{\boldsymbol{r}(t+\Delta t) - \boldsymbol{r}(t)}{1 - \overline{\boldsymbol{r}(t)}\boldsymbol{r}(t+\Delta t)}\right|\\ \amp =2\cdot \frac{\left|\frac{\boldsymbol{r}(t+\Delta t) - \boldsymbol{r}(t)}{\Delta t}\right|}{|1 - \overline{\boldsymbol{r}(t)}\boldsymbol{r}(t+\Delta t)|}\cdot | \Delta t |\text{.} \end{align*}
Now, as $\Delta t \to 0\text{,}$ the numerator in the above quotient goes to $|\boldsymbol{r}^\prime(t)|$ and the denominator goes to $1 - |\boldsymbol{r}(t)|^2\text{,}$ and we arrive at the following hyperbolic arc-length differential.
###### Definition5.3.6.
If $\boldsymbol{r}: [a,b] \to \mathbb{D}$ is a smooth curve in the hyperbolic plane, define the length of $\boldsymbol{r}$, denoted ${\cal L}(\boldsymbol{r})\text{,}$ to be
\begin{equation*} {\cal L}(\boldsymbol{r}) = \int_a^b \frac{2}{1 - |\boldsymbol{r}(t)|^2}~|\boldsymbol{r}^\prime(t)|dt\text{.} \end{equation*}
One can immediately check that the hyperbolic distance between two points in $\mathbb{D}$ corresponds to the length of the hyperbolic line segment connecting them.
The proof of this theorem is left as an exercise. One can prove that hyperbolic reflections preserve arc-length as well. This should come as no surprise, given the construction of the distance formula at the start of this section. Still, one can prove this fact from our definition of arc-length (Exercise 5.3.6). Thus, all hyperbolic reflections and all transformations in $\cal H$ are hyperbolic isometries: they preserve the hyperbolic distance between points in $\mathbb{D}\text{.}$
Another consequence of the invariance of distance, when applied to hyperbolic rotations, is the following:
Suppose $u$ and $v$ are on the same hyperbolic circle centered at $p\text{.}$ That is, these points are on the same type II cline with respect to $p$ and $p^*\text{,}$ so there exists a hyperbolic rotation that fixes $p$ and maps $u$ to $v\text{.}$ Thus, $d_H(p,u) = d_H(T(p),T(u)) = d_H(p,v)\text{.}$ It follows that any two points on the hyperbolic circle centered at $p$ are equidistant from $p\text{.}$
We are now in a position to argue that in the hyperbolic plane, the shortest path (geodesic) connecting two points $p$ and $q$ is along the hyperbolic line through them.
Proof Sketch: We first argue that the geodesic from 0 to a point $c$ on the positive real axis is the real axis itself.
Suppose $\boldsymbol{r}(t) = x(t) + iy(t)$ for $a \leq t \leq b\text{,}$ is an arbitrary smooth curve from 0 to $c$ (so $\boldsymbol{r}(a) = 0$ and $\boldsymbol{r}(b) = c$).
Suppose further that $x(t)$ is nondecreasing (if our path backtracks in the $x$ direction, we claim the path cannot possibly be a geodesic). Then
\begin{equation*} {\cal L}(\boldsymbol{r}) = \int_a^b\frac{2}{1-[(x(t))^2 + (y(t))^2]}\sqrt{(x^\prime(t))^2 + (y^\prime(t))^2}~dt. \end{equation*}
The hyperbolic line segment from 0 to $c$ can be parameterized by $\boldsymbol{r}_0(t) = x(t) + 0i$ for $a \leq t \leq b\text{,}$ which has length
\begin{equation*} {\cal L}(\boldsymbol{r}_0) = \int_a^b\frac{2}{1-[x(t)]^2}\sqrt{(x^\prime(t))^2}~dt\text{.} \end{equation*}
The curve $\boldsymbol{r}_0$ is essentially the shadow of $\boldsymbol{r}$ on the real axis.
One can compare the integrands directly to see that ${\cal L}(\boldsymbol{r}) \geq {\cal L}(\boldsymbol{r}_0)\text{.}$
Since transformations in ${\cal H}$ preserve arc-length and hyperbolic lines, it follows that the shortest path between any two points in $\mathbb{D}$ is along the hyperbolic line through them.
Recall our formula for the hyperbolic distance between two points in Theorem 5.3.3:
\begin{equation*} d_H(p,q) = \ln\left[\frac{|1-\overline{p}q|+|q-p|}{|1-\overline{p}q|-|q-p|}\right]. \end{equation*}
This expression is always non-negative because the quotient inside the natural log is always greater than or equal to 1. In fact, the expression equals 1 (so that the distance equals 0) if and only if $p = q\text{.}$
Note further that this formula is symmetric. Interchanging $p$ and $q$ leaves the distance unchanged.
Finally, the hyperbolic distance formula satisfies the triangle inequality because hyperbolic lines are geodesics.
###### Example5.3.11.Two paths from $\boldsymbol{p}$ to $\boldsymbol{q}$.
Two paths from $p = .5i$ to $q = .5+.5i$ are shown below: the (solid) hyperbolic segment from $p$ to $q\text{,}$ and the (dashed) path $\boldsymbol{r}$ that looks like a Euclidean segment. Which path is shorter?
We may compute the length of the hyperbolic segment connecting $p$ and $q$ with the distance formula from Theorem 5.3.3. This distance is approximately 1.49 units.
By contrast, consider the path in $\mathbb{D}$ corresponding to the Euclidean line segment from $p$ to $q\text{.}$ This path may be described by $\boldsymbol{r}(t) = t + \frac{1}{2}i$ for $0 \leq t \leq \frac{1}{2}\text{.}$ Then $\boldsymbol{r}^\prime(t) = 1$ and
\begin{align*} {\cal L}(\boldsymbol{r}) \amp = \int_0^{\frac{1}{2}}\frac{2}{1-(t^2+\frac{1}{4})}dt\\ \amp \approx 1.52\text{.} \end{align*}
It is no surprise that the hyperbolic segment connecting $p$ to $q$ is a shorter path in $(\mathbb{D},{\cal H})$ than the Euclidean line segment connecting them.
###### Example5.3.12.Perpendicular bisectors in $(\mathbb{D},{\cal H})$.
For any two points $p$ and $q$ in $\mathbb{D}\text{,}$ we may construct the perpendicular bisector to hyperbolic segment $pq$ by following the construction in Euclidean geometry. Construct both the hyperbolic circle centered at $p$ that goes through $q$ and the hyperbolic circle centered at $q$ that goes through $p\text{.}$ The hyperbolic line through the two points of intersection of these circles is the perpendicular bisector to segment $pq\text{,}$ labeled $L$ in the following diagram.
Hyperbolic reflection about $L$ maps $p$ to $q$ and $q$ to $p\text{.}$ Since hyperbolic reflections preserve hyperbolic distances, each point on $L$ is hyperbolic equidistant from $p$ and $q\text{.}$ That is, for each $z$ on $L\text{,}$ $d_H(z,p)=d_H(z,q)\text{.}$
In Euclidean geometry one uses perpendicular bisectors to construct the circle through three noncollinear points. This construction can break down in hyperbolic geometry. Consider the three points $p, q\text{,}$ and $r$ in Figure 5.3.13. The corresponding perpendicular bisectors do not intersect. There is no point in $\mathbb{D}$ hyperbolic equidistant from all three of these points. In particular, in hyperbolic geometry, there need not be a hyperbolic circle through three noncollinear points.
### ExercisesExercises
###### 1.
Suppose $0\lt x \lt 1$ and $L$ is a hyperbolic line about which $x$ gets inverted to the origin. (Such an inversion was constructed in Theorem 5.1.3.) For a real number $h\text{,}$ let $w$ be the image of $x+h$ under this inversion. Prove that $w = \frac{-h}{1-x^2-hx}\text{.}$
###### 2.
Determine a point in $\mathbb{D}$ whose hyperbolic distance from the origin is 2,003,007.4 units.
###### 3.
Suppose $L$ is any hyperbolic line, and $C$ is any cline through the ideal points of $L\text{.}$ For any point $z$ on $L\text{,}$ its perpendicular distance to $C$ is the length of the hyperbolic segment from $z$ to $C$ that meets $C$ at right angles. Prove that the perpendicular distance from $C$ to $L$ is the same at every point of $L\text{.}$ Hint: Use the fact that distance is an invariant of hyperbolic geometry.
###### 4.
Determine the hyperbolic distance from the point $p = 0.5$ to the point $q = 0.25 + 0.5i\text{.}$
###### 6.
Hyperbolic reflections preserve distance in $(\mathbb{D},{\cal H})$
1. Use the definition of arc-length to prove that hyperbolic reflection about the real axis preserves arc-length.
2. Use part (a) and Theorem 5.3.7 to argue that hyperbolic reflection about any hyperbolic line preserves arc-length in $(\mathbb{D}, {\cal H})\text{.}$
###### 7.
Suppose $z_0$ is in the hyperbolic plane and $r > 0\text{.}$ Prove that the set $C$ consisting of all points $z$ in $\mathbb{D}$ such that $d_H(z,z_0) = r$ is a Euclidean circle. | 6,331 | 19,314 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-18 | latest | en | 0.88587 |
http://slideplayer.com/slide/4052937/ | 1,508,279,632,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822513.17/warc/CC-MAIN-20171017215800-20171017235800-00412.warc.gz | 297,235,539 | 19,938 | # Mole-to-Mole Conversions
## Presentation on theme: "Mole-to-Mole Conversions"— Presentation transcript:
Mole-to-Mole Conversions
Chapter 8 :Quantities inChemical Reactions The study of the numerical relationship between chemical quantities in a chemical reaction is called reaction stoichiometry Mole-to-Mole Conversions the balanced equation is the “recipe” for a chemical reaction the equation 3 H2(g) + N2(g) 2 NH3(g) tells us that 3 molecules of H2 react with exactly 1 molecule of N2 and make exactly 2 molecules of NH3 or 3 molecules H2 1 molecule N2 2 molecules NH3 (in this reaction)
Example: Sodium chloride, NaCl, forms by the following reaction between sodium and chlorine. How many moles of NaCl result from the complete reaction of 3.4 mol of Cl2? Assume there is more than enough Na. 2 Na(s) + Cl2(g) 2 NaCl(s)
Check the Solution: 3.4 mol Cl2 6.8 mol NaCl
Example: How many moles of NaCl result from the complete reaction of 3.4 mol of Cl2 in the reaction below? 2 Na(s) + Cl2(g) 2 NaCl(s) Information Given: 3.4 mol Cl2 Find: ? moles NaCl CF: 1 mol Cl2 2 mol NaCl SM: mol Cl2 mol NaCl Check the Solution: 3.4 mol Cl2 6.8 mol NaCl The units of the answer, moles NaCl, are correct. The magnitude of the answer makes sense since the equation tells us you make twice as many moles of NaCl as the moles of Cl2 .
Mass-to-Mass Conversions
we know there is a relationship between the mass and number of moles of a chemical 1 mole = Molar Mass in grams the molar mass of the chemicals in the reaction and the balanced chemical equation allow us to convert from the amount of any chemical in the reaction to the amount of any other
Example: In photosynthesis, plants convert carbon dioxide and water into glucose, (C6H12O6), according to the following reaction. How many grams of glucose can be synthesized from 58.5 g of CO2? Assume there is more than enough water to react with all the CO2.
Example: How many grams of glucose can be synthesized from 58
Example: How many grams of glucose can be synthesized from 58.5 g of CO2 in the reaction? 6 CO2(g) + 6 H2O(l) 6 O2(g) + C6H12O6(aq) Information Given: 58.5 g CO2 Find: g C6H12O6 CF: 1 mol C6H12O6 = g 1 mol CO2 = g 1 mol C6H12O6 6 mol CO2 Write a Solution Map: g CO2 mol CO2 mol C6H12O6 g C6H12O6
Check the Solution: 58.5 g CO2 = 39.9 g C6H12O6
Example: How many grams of glucose can be synthesized from 58.5 g of CO2 in the reaction? 6 CO2(g) + 6 H2O(l) 6 O2(g) + C6H12O6(aq) Information Given: 58.5 g CO2 Find: g C6H12O6 CF: 1 mol C6H12O6 = g 1 mol CO2 = g 1 mol C6H12O6 6 mol CO2 SM: g CO2 mol CO2 mol C6H12O6 g C6H12O6 Check the Solution: 58.5 g CO2 = 39.9 g C6H12O6 The units of the answer, g C6H12O6 , are correct. It is hard to judge the magnitude.
Limiting reactant and Yield
Limiting Reagent (or reactant): The reactant that is completely consumed in a chemical reaction. Theoretical yield: The amount of product that can be made in a chemical reaction based on the amount of limiting reactant. Actual yield: The amount of product actually produced by a chemical reaction. Because of both controllable and uncontrollable factors, the actual yield of product will always be less than the theoretical yield.
Tro's Introductory Chemistry, Chapter 8
Example: When 11.5 g of C are allowed to react with g of Cu2O in the reaction below, 87.4 g of Cu are obtained. Find the Limiting Reactant, Theoretical Yield and Percent Yield. Tro's Introductory Chemistry, Chapter 8
Limiting Reactant = Cu2O Theoretical Yield = 101.7 g
Example: When 11.5 g of C reacts with g of Cu2O, 87.4 g of Cu are obtained. Find the Limiting Reactant, Theoretical Yield and Percent Yield. Cu2O(s) + C(s) 2 Cu(s) + CO(g) Information Given: 11.5 g C, g Cu2O 87.4 g Cu produced Find: Lim. Rct., Theor. Yld., % Yld. CF: 1 mol C = g; 1 mol Cu = g; 1 mol Cu2O = g; mol Cu2O 2 mol Cu; 1 mol C 2 mol Cu Check the Solutions: Limiting Reactant = Cu2O Theoretical Yield = g Percent Yield = 85.9% The Percent Yield makes sense as it is less than 100%. Tro's Introductory Chemistry, Chapter 8 | 1,256 | 4,066 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2017-43 | longest | en | 0.89658 |
http://www.glencoe.com/sec/math/studytools/cgi-bin/msgQuiz.php4?isbn=0-07-860861-9&chapter=2&headerFile=4&state=va | 1,386,742,356,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164032243/warc/CC-MAIN-20131204133352-00020-ip-10-33-133-15.ec2.internal.warc.gz | 356,128,290 | 5,436 | 1. The cost of producing an item is \$5 per item plus an initial cost of \$2000. The selling price is \$10 per item. Find the break-even point. A. 4500 items B. 4000 items C. 450 items D. 400 items Hint 2. Solve the system of equations y = 0.5x and 4y = x - 2 by graphing. A. B. C. D. Hint 3. If you solve the following system of equations by elimination, which of the following is the best choice for the first step?2x + y - z = 3x + y + z = 5x - 2y + z = 2 A. Both methods will work. B. Add the first and second equations to eliminate the z variable. C. Neither method will work. D. Subtract the second and third equation to eliminate the z variable. Hint 4. Solve the system of equations by substitution.x = zx - 2y + z = 62x + y - 2z = 1What is the value of x + y + z? A. 10 B. 8 C. 11 D. 9 Hint 5. A. B. C. D. Hint 6. Find the image of after Rot90 · Ry-axis if the vertices areA(2, -3), B(6, -3), and C(2, 5). A. A'(3, -2), B'(3, -6), C'(-5, -2) B. A'(3, 2), B'(3, 6), C'(5, -2) C. A'(-3, -2), B'(-3, -6), C'(-5, -2) D. A'(-3, 2), B'(-3, 6), C'(5, 2) Hint 7. Which of the following is a graph of the inequalities l + g > 100 and l > 70? A. B. C. D. Hint 8. Use the function P(x, y) = 40x + 60y to determine how many of each item should be produced in order to maximize profit. A. (300, 500) B. (100, 800) C. (100, 400) D. (500, 400) Hint 9. A. impossible B. C. D. Hint 10. Use matrices to determine the coordinates of the image of with vertices A(-3,4), B(-5,2) and C(-6,5) once it is rotated 90°. A. A'(3,-4), B(5,-2) and C'(6,-5) B. A'(3,4), B(5,2) and C'(6,5) C. A'(-4,-3), B(-2,-5) and C'(-5,-6) D. A'(-3,-4), B(-5,-2) and C'(-6,-5) Hint 11. Find the inverse of . A. 0 B. does not exist C. D. Hint 12. Find the inverse of . A. B. C. D. Hint 13. Solve the system of inequalities by graphing.x + y 42x - y < 4y 0 A. B. C. D. Hint 14. Find the maximum value of f(x, y) = 2x + y - 4 for the system of inequalities:y -3x + 1y x - 4x 0y 0 A. 2 B. unbounded C. alternate optimal solutions D. infeasible Hint | 788 | 2,009 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2013-48 | latest | en | 0.718826 |
https://www.cfd-online.com/Forums/main/9231-problem-solving-such-equation.html | 1,503,063,673,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104636.62/warc/CC-MAIN-20170818121545-20170818141545-00685.warc.gz | 865,229,058 | 16,312 | # problem in solving such an equation
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May 26, 2005, 11:44 problem in solving such an equation #1 R Guest Posts: n/a Sponsored Links Hi, I have problem in solving such an equation dy/dt=u(t)-S(r)*[1+(dy/dr)^2] where u(t)=U[1+Vsin(w*t)] We are solving for y(r) and U is an constant =37, V >1. S(r)=S*(r-a)^(1/3) and S is also constant =3.33. Follwing time depedent boundary conditions is used. dy/dt=0 if u(t)>=S(a) dy/dr=0 if u(t)<=S(a) I do not have problem when I solve this equations when V<1. Also I do not have problem when S(r)=S even if V>1. Please help me thanks
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May 27, 2005, 00:48 Re: problem in solving such an equation #2 harish Guest Posts: n/a What kind of numerical scheme are you using to solve the problem ? -Harish
May 27, 2005, 03:25 Re: problem in solving such an equation #3 R Guest Posts: n/a Actually I have used CN, Fully implicit and expilcit with back differencing and central differencing schemes.
May 27, 2005, 05:01 Re: problem in solving such an equation #4 Tom Guest Posts: n/a Have you tried to write down the analytic solution - your pde is only first order so you can write down a general expression for the characteristics - from which you should see the source of your problem! Also isn't S(a)=0 your boundary condition so that dy/dt = 0 on r=a (I assame r=a is the boundary from your notation).
May 27, 2005, 07:04 Re: problem in solving such an equation #5 R Guest Posts: n/a I have tried the analytic solutions of such equation. The analytic solution is only possible for the case of V<<1. But the case of V>1 analytical solution is wrong. Can you give me examples how can analytical solutions helps in such case. Yes you are write about the boundary condition. R
May 27, 2005, 07:49 Re: problem in solving such an equation #6 Tom Guest Posts: n/a I'm not sure what you mean by the analytic solution being wrong - unless you mean you tried to obtain a series solution in V<<1? Otherwise the (correct) analytic solution cannot be wrong since it is the solution to the equation! What I meant by anaytic solution was that you can reduce your pde, using the method of characteristics, to a set of odes for dt/ds, dr/ds, dy/ds dp/ds and dq/ds where p=dy/dt, q=dy/dr and s is the variable along the characteristic. The analytic solution helps for two reasons: (1) It gives something to test your numerics against and (2) If there is a problem with the existence of the solution (i.e. finite time blowup) then you will be able to see why and in what parameter regions it happens.
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Contact Us - CFD Online - Top | 946 | 3,459 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2017-34 | longest | en | 0.936957 |
https://stats.stackexchange.com/questions/546985/why-sumc-c-neq-cpcx1-pcxp1-p-leq-c-1-frac1-pc-11 | 1,660,580,275,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572192.79/warc/CC-MAIN-20220815145459-20220815175459-00164.warc.gz | 487,677,802 | 67,433 | # Why $\sum^C_{c\neq c^*}P(c|x)[1-P(c|x)]+p^*(1-p^*) \leq (C-1)\frac{1-p^*}{C-1}[1-\frac{1-p^*}{C-1}]+p^*(1-p^*)?$
I have the following theorem in my textbook:
As the number of samples goes to infinity the error rate of 1-NN is no more than twice the Bayes error rate.
Proof sketch:
Abbreviate notation $$P(c|x) := P(Y=c|x)$$
The expected (Bayes) error of the Bayes classifier (at x) is $$1-\text{ max }_{c\in\{1,...,C\}}P(c|x)$$
and the expected rate of $$1-NN$$ (at x) is
$$\sum^C_{c=1}P(c|x_{nn})[1-P(c|x)]$$
observe that as the number of samples goes to infinity, $$m\to \infty$$,
$$P(c|x)\approx P(c|x_{nn})$$
thus the expected rate of 1-NN (at x) is $$\sum^C_{c=1}P(c|x)[1-P(c|x)]$$
We need to show $$\sum^C_{c=1}P(c|x)[1-P(c|x)]\leq 2[1-\text{ max }_{c\in\{1,...,C\}}P(c|x)]$$:
Let $$c^*=\text{ argmax }_{c\in\{1,...,C\}}P(c|x)$$ and $$p^*=P(c^*|x)$$. Observe that $$\sum^C_{c=1}P(c|x)[1-P(c|x)]=\sum^C_{c\neq c^*}P(c|x)[1-P(c|x)]+p^*(1-p^*)\\ \leq (C-1)\frac{1-p^*}{C-1}[1-\frac{1-p^*}{C-1}]+p^*(1-p^*)\\ = (1-p^*)[1-\frac{1-p^*}{C-1}+p^*]$$
Where the second line follows since the sum is maximised when all "$$P(c|x)$$" have the same value. And since $$p^*<1$$ we are done.
I don't completely understand how the second line follows. Could someone show me please?
• Are you sure your expression for $c^* = \arg \max \dots$ is correct? Oct 4, 2021 at 18:19
## 2 Answers
Let's define $$y_c:=P(c|x)$$, $$\forall{c} \in \{1,2,\ldots, C\}$$, then we have $$y_{c^{*}}=p^{*}$$
Also, let $$f\left(y_1,y_2,\ldots,y_{c^{*}-1},y_{c^{*}+1},\ldots,y_C\right)=\sum\limits_{c\neq c^{*}}y_c(1-y_c)$$, this is the function we want to maximize (note the absence of $$c^{*}$$ in the variables).
Also, we have a constraint that $$\sum\limits_{c=1}^{C}y_c=1$$, since the posterior probabilities must sum to $$1$$.
Let's define the constraint function $$g\left(y_1,y_2,\ldots,y_{c^{*}-1},y_{c^{*}+1},\ldots,y_C\right)=\sum\limits_{c\neq c^{*}}y_c + p^{*}-1=0$$,
s.t. we have to maximize $$f$$ w.r.t. the constraint $$g$$.
It means we must have $$\nabla f = \lambda \nabla g$$ for some $$\lambda$$, when $$f$$ is an extremum (using the Lagrange multiplier).
\implies \begin{align} \begin{bmatrix} 1-2y_1 \\ 1-2y_2 \\ \vdots \\ 1-2y_{c^{*}-1} \\ 1-2y_{c^{*}+1} \\ \vdots \\ 1-2y_{C} \end{bmatrix} \end{align} =\lambda \begin{align} \begin{bmatrix} 1 \\ 1 \\ \vdots \\ 1 \\ 1 \\ \vdots \\ 1 \end{bmatrix} \end{align} $$\implies y_c = \dfrac{\lambda-1}{2}$$, $$\forall{c} \neq c^{*}$$
Also, we must have the constraint satisfied, i.e.,
$$\sum\limits_{c\neq c^{*}}y_c + p^{*}-1=\sum\limits_{c\neq c^{*}}\dfrac{\lambda-1}{2}+p^{*}-1= 0$$
$$\implies (C-1)\dfrac{\lambda-1}{2}+p^{*}-1= 0$$
$$\implies y_c=\dfrac{\lambda-1}{2}=\dfrac{1-p^{*}}{C-1}$$, $$\forall{c} \neq c^{*}$$
Hence, at maximum, we have,
$$f_{max}\left(y_1,y_2,\ldots,y_{c^{*}-1},y_{c^{*}+1},\ldots,y_C\right)=\sum\limits_{c\neq c^{*}}y_c(1-y_c)=\sum\limits_{c\neq c^{*}}\dfrac{1-p^{*}}{C-1}\left(1-\dfrac{1-p^{*}}{C-1}\right)$$
$$=(C-1).\dfrac{1-p^{*}}{C-1}\left(1-\dfrac{1-p^{*}}{C-1}\right)$$
Now, since we have $$f \leq f_{max}$$,
$$\implies \sum\limits^C_{c\neq c^*}P(c|x)[1-P(c|x)] \leq (C-1)\frac{1-p^*}{C-1}[1-\frac{1-p^*}{C-1}]$$
$$\implies \sum\limits^C_{c\neq c^*}P(c|x)[1-P(c|x)]+p^*(1-p^*) \leq (C-1)\frac{1-p^*}{C-1}[1-\frac{1-p^*}{C-1}]+p^*(1-p^*)$$
• thanks for the elaborate answer! Oct 5, 2021 at 8:59
Sorry, quick answer since I'm on my way out the door: you need to think about maximizing the curve
$$\sum_{c\neq c^*} x_c(1-x_c)\,,\hspace{3em} \text{with } x_c\geq0\,,\hspace{3em}\text{and}\,\, \sum_{c\neq c*} = 1-x_{c^*}\,.$$
A bit of straightforward calculus (or noticing that $$x(1-x) in the domain) will show you that this is maximized when $$x_1=x_2=\ldots=x_{C}$$, here when $$x_c = (1-x_{c^*})/(C-1)$$. So
$$\sum_{c-c^*} P(c|x)[1-P(c|x)] \leq \sum_{c-c^*} P(c^*|x)[1-P(c^*|x)] = (C-1)P(c^*|x)[1-P(c^*|x)] = (C-1)\left(\frac{1-p^*}{C-1}\right)\left[1-\frac{1-p^*}{C-1}\right]\,.$$
The result follows.
• How exactly did you know that it's maximized when $x_1=x_2=...=x_C$? Also, why exactly is $P(c^*|x)=\frac{1-p^*}{C-1}$? Isn't $P(c^*|x)=p^*$? Also, don't understand why $x_c=1(1-x_{c^*})/(C-1)$. Could you please elaborate on your answer? Oct 4, 2021 at 17:59
• 1. The straight forward calculus is Lagrange Multiplies: to maximize $f(x)$ subject to $g(x)=c$, find the joint solution to $\nabla f(x) = \lambda \nabla g(x)$ and $g(x) = c$. This is pretty straight forward in this case if you know calculus, but if you don't writing out the computation isn't going to be fairly useful. I would give it a try yourself and ask if you have questions. 2. $P(c^*|x) \neq (1-p*)/(C-1)$, instead we're trying to find the maximum that the left hand side of my expression could possibly be to bound it. By (1), that maximum occurs when the probabilities are equal.
– Nate
Oct 4, 2021 at 18:27
• 3. If all $x_c$ are equal for $c\neq c^*$ then $1-p^* = \sum_{c\neq c^*} x_c = (1-C)x_c$, so $x_c = (1-p^*)/(1-C)$.
– Nate
Oct 4, 2021 at 18:37 | 2,187 | 5,039 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 44, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2022-33 | longest | en | 0.710566 |
https://studylib.net/doc/25709757/lesson-5-full | 1,685,825,393,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649343.34/warc/CC-MAIN-20230603201228-20230603231228-00052.warc.gz | 594,391,675 | 15,812 | # Lesson-5-full
```Adaptation and Behavior
Time Frame: 120 minutes (split into two lessons)
Learning Standards:
Science
Life Science: Plant (and Animal) Structures and Functions
1) Give examples of how inherited characteristics may change over time as
adaptations to changes in the environment that enable organisms to survive, e.g.,
shape of beak or feet, placement of eyes on head, length of neck, shape of teeth,
color.
Skills of Inquiry
1) Ask questions and make predictions that can be tested.
2) Select and use appropriate tools and technology in order to extend observations.
3) Keep accurate records while conducting simple investigations or experiments.
4) Conduct multiple trials to test a prediction. Compare the result of an investigation
or experiment with the prediction.
5) Recognize simple patterns in data and use data to create a reasonable
explanation for the results of an investigation of experiment.
6) Record data and communicate findings to others using graphs, charts, maps,
models, and oral and written reports.
Math
Data Analysis, Statistics, and Probability
1) 5.D.2 Construct and interpret line plots, line graphs, and bar graphs. Interpret
and label circle graphs.
Number Sense and Operations
1) 6.N.9 Select and use appropriate operations to solve problems involving addition,
subtraction, multiplication, division, and positive integer exponents with whole
numbers, and with positive fractions, mixed numbers, decimals, and percents.
English / Language Arts
Oral Presentation
1) 3.8 Give oral presentations for various purposes, showing appropriate changes in
delivery and using language for dramatic effect.
Student will be able to:
1) Predict what type of food different birds eat based on the shape of their beak.
2) Experiment with model beaks and food sources and collect, record, graph, and
analyze data to determine which type of beak is best suited for which type of
food.
3) Explain how birds are adapted to eat a certain type of food and describe what
would happen to a bird population if its environment changed.
4) Present the results of the bird beak experiment to the class.
Resources and Materials for both parts:
Item
Science notebooks
Bird beak overhead (pictures of bird
beaks)
data and analysis, and graph
Bird beak challenge instructions
Stopwatches or clock
Demonstration materials
Box of food coloring
Extras for the different challenges
Challenge #1:
Challenge #1 instructions
Cup with a marked line
Shoestring
Dropper
Sponge strip
Challenge #2:
Challenge #2 instructions
Gummy worms
Cup
Straw
Chopsticks
Wrench/Pliers
Potting soil
Deep metal pan
Challenge #3:
Challenge #3 instructions
Sunflower seeds
Cup
Pliers
Chopsticks
Tweezers
Challenge #4:
Challenge #4 instructions
Styrofoam pieces
Cup
Chopsticks
Tweezers
Slotted spoon
Shallow pan
Amount
(in binder)
(in bin)
(in binder)
6 (in bin or from classroom)
(in bin)
1
1
(use materials from the challenges for the
other demonstrations)
Various amounts
(in bin)
10
1
1
1
1
1
(in bin)
10
1 bag
1
2
1
1
1 bag
1
(in bin)
10
1 bag
1
1
1 pair
1
(in bin)
10
1 bag
1
1 pair
1
1
1
Challenge #5:
Challenge #5 instructions
Rice
Circular piece of Styrofoam
Dropper
Tongs
Tweezers
Challenge #6:
Challenge #6 instructions
Marshmallows
String
Chopsticks
Tongs
Skewers
Build-a-bird activity cutouts
(in bin)
10
1 bag
1
1
1
1
(in bin)
10
1 bag
1 roll
1 pair
1
1
(in bin)
General Teacher Background Information
Animal adaptations are any body shape, process, or behavior that allows an organism
to survive in its environment. Populations of animals change over time in response to
their environment. Why do birds have different shaped beaks? Birds have many
different kinds of beaks, depending on what they eat and where their food source is. For
instance, birds may find their food in water, mud, flowers, seeds, or in wood. The
different shapes of beaks allow easier access to these various food supplies. If an
environment altered, organisms within the area might not survive. Adaptation to a
particular environment occurs over time as organisms best suited to the environment
survive and reproduce, and pass their inherited traits to the next generation.
Lesson 5 Part I: Beaks and Food
Focus Activity: Use an overhead (or PowerPoint) with pictures of birds and their beaks
(the beaks of these birds reflect the supplies you are using). Have students open their
science notebooks and predict what each bird eats with its beak and write any ideas
they have about how it might use its beak to eat. Tell students to sketch the beak next
to their predictions.
Introduction: Pass out the Bird Adaptation handouts to each student. Review the
definition of adaptation from the last science lesson. Discuss the following questions as
a class. What does adaptation mean? What are some examples of adaptations? How
must be inherited.)
Then, relate adaptations to birds. Ask the students to share with the class what they
know about birds. What makes a bird a bird? What do birds need to survive? What
kinds of food do they think birds eat? (Insects, seeds, berries, and meat are among the
most common.) Where do birds live?
Activity:
1) Arrange the following materials in front of the class (set this up before class and
use these same materials for the challenge stations):
1. A tall, thin vase filled with colored water (replace the colored water the
clear water for the challenge station)
2. A dish of potting soil with gummy worms buried throughout
3. Sunflower seeds spread throughout a pan
4. A dish of water with styrofoam cubes floating in shallow water
5. Rice grains tucked into Styrofoam
6. Marshmallows hanging on strings
Tell students that each of these items represents a type of food eaten by various
birds. Ask students if they can hypothesize what a bird would have to do in order to
reach the food supply. Does the shape of a bird's beak limit what they can eat?
Teacher Information: The specific birds that eat each type of food should be
discussed at the end of class after the group presentations.
1. Nectar (colored water) needs to be sucked out Hummingbird
2. Worms (gummy worms) need to be dug and pulled out Robin
3. Seeds (sunflower seeds) need to be cracked open Sparrows, Finches
4. Fish (styrofoam pieces) need to be scooped out of the water Heron
5. Small insects (rice) need to be pried out of small crevices Woodpeckers
6. Meat (marshmallows) will need to be pulled off of bones Owls, Hawks
2) Divide Students into groups of three or more (there will be six different group
challenges, but you may want to have an extra station open for groups that work
faster, so divide students accordingly).
3) Make sure the challenge stations and equipment are set up around the room.
(Each station will have the challenge instructions, a food source, and a set of
three different utensils, which they are to use as sample "beaks.")
4) After reading their card, ask students to write their challenge number and which
"beak" they predict will work best for "eating" their specific "food" in their science
notebooks.
5) Each group will time in seconds how long it takes to get a certain amount of
"food" with each utensil. (See challenge cards). Students may work in groups to
complete one, some, or all of the challenges depending on how many days spent
on this activity.
Assign Roles: This activity will go more smoothly if each student has a different
role for each trial. If the students are in groups of three, each one can perform a
different task for each of the three trials. It may be helpful to write the
responsibilities for each role on the board so that students may refer to them
throughout the activity.
Materials Manager and Timer: set up the food source for the next beak,
and use the stopwatch or clock to time the trials
Reader and Recorder: Read the directions and record the data in the
chart provided
Experimenter: Conduct one trial for the bird beak experiment - use each
of the three beaks to get the specified amount of food
6) When the challenge card activities are done ask each group to calculate the
average time for each beak to eat their food source and make a bar graph to
represent their data.
Closure: Discuss the following ideas and questions as a class or in their groups. In
your challenge, some tools were better at getting food than others. How are the trial
tools you used relate to the different shapes of bird beaks? What did your challenge
Lesson 5 Part II: Beaks and Food
Focus Activity: Ask students to get back into last week’s groups. Ask the groups to
answer the analysis questions and assess their initial predictions in their lab notebooks.
Direct each group to make a brief presentation to the class about one of their
challenges (try to have each group present a different challenge). Groups should
describe their food model beaks and tell which beak worked the best. Then, they
should explain which shape of beak and bird from the overhead they think would best
suit their food source. Ask the class if they can think of any other adaptations that might
help each bird better survive in its habitat.
Class Presentation:
1) As a group, describe your food source and which type of beak was fastest at eating
this food (show the bar graph).
2) Then, explain which shape of beak and bird from the overhead would be best at
eating this type of food. Use the evidence from your challenge to back up your
explanations.
Bird Beak Analysis:
Complete the following questions in your science notebooks.
1) Write an explanation next to your prediction. Was your prediction correct or
incorrect? How did your ideas change after doing the experiment?
2) Describe what might happen to a bird population if its natural environment
experienced a natural disaster where all plants and animals were wiped out. For
example, what would happen to a woodpecker if all of the trees burned down?
What would happen to a heron if a farmer used a dangerous pesticide on his crops?
Activity:
1) In their groups, have students create a bird that feeds on a particular food source.
Be sure to include adaptations in addition to their beaks, such as long legs for a
wading bird like the heron. Draw a picture (or use the bird cutouts) and write a
description of what adaptations this bird has that allow it to successfully feed on its
food source.
2) Have the groups present their birds. Describe the bird and why it has certain
Closure: Discuss the following ideas and questions as a class. How is a heron better
adapted to catch a fish than a woodpecker? How is a duck better adapted to an aquatic
environment than an owl? How are different types of birds adapted to their
environments and food source? What would happen to a bird population if its food
source disappeared? What would happen to the hummingbirds that summer in
Williamstown if all the flowers died?
Assessment: Science notebooks responses, participation in discussion and lab
activities, bird beak graph and analysis questions, bird beak presentations
Challenge #1
Food source: graduated cylinder with water
Sample beaks: shoestring, dropper, sponge strip
Procedure:
1) Find out how many seconds it takes each beak to fill the
cup with water from the graduated cylinder.
2) Complete 3 trials with each beak.
3) Record your times in the data table.
4) Calculate the average time for each beak.
5) Construct a bar graph of the averages.
Challenge #2
Food source: gummy worms
Sample beaks: straw, chopsticks, wrench/pliers
Procedure:
1) Find out how many seconds it takes to remove three
gummy worms from the dirt using each beak.
2) Complete 3 trials with each beak, burying the worms
after each trial.
3) Record your times in the data table.
4) Calculate the average time for each beak.
5) Construct a bar graph of the averages.
Challenge #3
Food source: sunflower seeds
Sample beaks: pliers, chopsticks, tweezers
Procedure:
1) Find out how many seconds it takes each beak to crack
three sunflower seeds and remove the seeds inside.
2) Complete 3 trials with each beak.
3) Record your times in the data table.
4) Calculate the average time for each beak.
5) Construct a bar graph of the averages.
Challenge #4
Food source: Floating Styrofoam squares
Sample beaks: chopsticks, tweezers, slotted spoon
Procedure:
1) Find out how many seconds it takes each beak to
remove all of the Styrofoam squares from the water.
2) Complete 3 trials with each beak, returning the squares
after each trial.
3) Record your times in the data table.
4) Calculate the average time for each beak.
5) Construct a bar graph of the averages.
Challenge #5
Food source: rice (in Styrofoam)
Sample beaks: dropper, plastic tongs, tweezers
Procedure:
1) Find out how many seconds it takes for each beak to
remove twenty grains of rice from the bark of a tree
(Styrofoam).
2) Complete 3 trials with each beak, returning the rice to
the bark each time.
3) Record your times in the data table.
4) Calculate the average time for each beak.
5) Construct a bar graph of the averages.
Challenge #6
Food source: hanging marshmallows
Sample beaks: chopsticks, metal tongs, skewer
Procedure:
1) Find out how many seconds it takes for each beak to
remove five marshmallows from the strings.
2) Complete 3 trials with each beak.
3) Record your times in the data table.
4) Calculate the average time for each beak.
5) Construct a bar graph of the averages.
Bird Adaptations: Beaks and Food Part I
Predictions: Observe and interpret the 6 pictures provided of the different types of
birds. Open your science notebooks and predict what each bird eats with its beak and
write any ideas you have about how it might use its beak to eat. Sketch the beak next
too each prediction.
Procedure:
1) After you are in a group, go to a challenge station. Each station will have a different
food source and a set of three different utensils, which you are to use as sample
beaks.
3) Write your challenge number and which model beak you predict will work best for
eating the food in your science notebooks.
4) Each person in the group will have a different role for each trial.
Materials Manager and Timer: Set up the food source for the next beak and
use a stopwatch or clock to time the trials
Reader and Recorder: Read the directions and record the data in the chart
provided
Experimenter: Conduct one trial with each bird beak
5) Each group will time in seconds how long it takes to get a certain amount of food
with each model beak. (See challenge card.)
Bird Adaptations: Beaks and Food Part II
Class Presentation:
3) As a group, describe your food source and which type of beak was fastest at eating
this food (show the bar graph).
4) Then, explain which shape of beak and bird from the overhead would be best at
eating this type of food. Use the evidence from your challenge to back up your
explanations.
Bird Beak Analysis:
Complete the following questions in your science notebooks.
3) Write an explanation next to your prediction. Was your prediction correct or
incorrect? How did your ideas change after doing the experiment?
4) Describe what might happen to a bird population if its natural environment
experienced a natural disaster where all plants and animals were wiped out. For
example, what would happen to a woodpecker if all of the trees burned down?
What would happen to a heron if a farmer used a dangerous pesticide on his crops?
Extension Activity:
On another sheet of paper, create a bird that feeds on a particular food source. Be sure
to include adaptations in addition to their beaks, such as long legs for a wading bird like
the heron. Draw a picture (or use the bird cutouts) and write a description of what
adaptations this bird has that allow it to successfully feed on its food source.
Student Data and Analysis Sheet
Group Member Names: _________________________________________________
Challenge #____
Bird Beak Data Table
Type of Beak
Trial #1
Tested:
(in seconds)
Food source: _______________
Trial #2
( in seconds)
Trial #3
(in seconds)
To calculate the average time:
Trial #1 + Trial #2 + Trial #3
3 (number of trials)
=
Average time
Add the total number of seconds from each of the 3 trials.
Then, divide that number by 3.
This is the average time it takes the type of beak tested to get the food.
Average Time
(in seconds)
Build A Bird
Body | 3,826 | 16,276 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2023-23 | longest | en | 0.896523 |
https://discuss.elastic.co/t/aggregation-with-unique-count-and-sum-how-to-do-it/234422 | 1,675,111,385,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499829.29/warc/CC-MAIN-20230130201044-20230130231044-00667.warc.gz | 231,954,676 | 7,272 | # Aggregation with unique count and sum: how to do it?
Hello.
We have some documents in ES (7.7.0) that represent some kind of ecommerce sales transactions, that can be described as the following table
``````| tx_id | item | cost | price |
-------------------------------------
| 1 | item1 | 10 | 15 |
| 2 | item2 | 15 | 20 |
| 3 | item3 | 20 | 30 |
| 1 | item1 | 10 | 15 |
| 4 | item4 | 20 | 25 |
| 5 | item2 | 15 | 20 |
| 1 | item1 | 10 | 15 |
| 6 | item2 | 15 | 20 |
| 2 | item2 | 15 | 20 |
| 7 | item1 | 10 | 15 |
``````
For reasons that are too long to explains, some of the documents contains "duplicated transactions".
In this example data table, we have 3 documents with the same transaction_id=1 and 2 documents with the same transaction_id=2.
We can't change the application that writes the transactions in the documents to avoid the duplication, so we have to live with that constraint.
What we are trying to achieve, is a consolidated report with the following data:
``````| item | unit_cost | unit_price | sold units | cost | revenue |
----------------------------------------------------------------
| item1 | 10 | 15 | 2 | 20 | 30 | <-- tx_id 1, 7, but tx_id=1 appears 3 times in the data table
| item2 | 15 | 20 | 3 | 45 | 60 | <-- tx_id 2, 5, 6, but tx_id=2 appears 2 times in the data table
| item3 | 20 | 30 | 1 | 20 | 30 | <-- tx_id 3
| item4 | 20 | 25 | 1 | 20 | 25 | <-- tx_id 4
``````
We are struggling to get this aggregated table in kibana.
We tried to create `terms` buckets for the fields `item`, `cost` and `price`, and calculating the metrics `unique count of tx_id` to get the `sold_units`, but then how to calculate the metrics for the `cost` and `revenue`?
Using the `metric sum` will not work, because the sum of the `bucket item:item1 --> unit_cost:10 --> unit_price:15` contains 4 documents, because of the duplication of tx_id=1
Any hints or suggestions?
Thanks.
SLL
Hey @simonlucalandi! I'm not 100% certain if Kibana can produce something exactly like that but I think we can get close...
Can you can create a terms aggregation on your `tx_id`? Just trying to follow along to this previous issue I found: Sum of unique ids
Hello!
Thank you for the suggestion.
I'm now trying to use the "transform data" feature to create a pivot that removes the duplicated transaction (using "Max" aggregation) that save the clean data in a new index, and then create visualization on this new index.
It seams to be fit our needs.
This topic was automatically closed 28 days after the last reply. New replies are no longer allowed. | 818 | 2,843 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2023-06 | latest | en | 0.781912 |
https://questioncove.com/updates/524c1ecbe4b0e0b65598e907 | 1,627,818,264,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154175.76/warc/CC-MAIN-20210801092716-20210801122716-00197.warc.gz | 484,061,006 | 5,365 | Mathematics
OpenStudy (anonymous):
How are the functions y = x and y = x – 3 related? How are their graphs related? Can someone explain this to me
OpenStudy (anonymous):
Each output for y = x – 3 is 3 more than the corresponding output for y = x. The graph of y = x – 3 is the graph of y = x translated down 3 units. Each output for y = x – 3 is 3 more than the corresponding output for y = x. The graph of y = x – 3 is the graph of y = x translated up 3 units. Each output for y = x – 3 is 3 less than the corresponding output for y = x. The graph of y = x – 3 is the graph of y = x translated down 3 units. Each output for y = x – 3 is 3 less than the corresponding output for y = x. The graph of y = x – 3 is the graph of y = x translated up 3 units.
OpenStudy (anonymous):
y = x and y = x – 3 are these related do to same slope
OpenStudy (anonymous):
They do have the same slope just different y's the first one passes at (0,0)and the second at (0,-3)
OpenStudy (zaarin):
|dw:1380725762921:dw|
OpenStudy (anonymous):
that would be -x+3 wouldn't it
OpenStudy (anonymous):
@RavenLynette what your saying is that c is the right answer Each output for y = x – 3 is 3 less than the corresponding output for y = x. The graph of y = x – 3 is the graph of y = x translated down 3 units. "y's the first one passes at (0,0)and the second at (0,-3)" Right?
OpenStudy (anonymous):
Right :) | 405 | 1,396 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2021-31 | latest | en | 0.933347 |
https://engineeringtoolbox.com/stress-restricting-thermal-expansion-d_1756.html | 1,720,930,072,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514548.45/warc/CC-MAIN-20240714032952-20240714062952-00779.warc.gz | 201,496,628 | 11,532 | Engineering ToolBox - Resources, Tools and Basic Information for Engineering and Design of Technical Applications!
# Restricted Thermal Expansion - Force and Stress
## Stress and force when thermal expansion a pipe, beam or similar is restricted.
Linear expansion due to change in temperature can be expressed as
dl = α lo dt (1)
where
dl = elongation (m, in)
α = temperature expansion coefficient (m/mK, in/in oF)
lo = initial length (m, in)
dt = temperature difference (oC, oF)
The strain - or deformation - for an unrestricted expansion can be expressed as
ε = dl / lo (2)
where
ε = strain - deformation
The Elastic modulus (Young's Modulus) can be expressed as
E = σ / ε (3)
where
E = Young's Modulus (Pa (N/m2), psi)
σ = stress (Pa (N/m2), psi)
### Thermal Stress
When restricted expansion is "converted" to stress - then (1), (2) and (3) can be combined to
σdt = E ε
= E dl / lo
= E α lo dt / lo
= E α dt (4)
where
σdt = stress due to change in temperature (Pa (N/m2), psi)
### Axial Force
The axial force acted by the restricted bar due to change in temperature can be expressed as
F = σdt
= E α dt A (5)
where
F = axial force (N)
A = cross-sectional area of bar (m2, in2)
#### Example - Heated Steel Pipe - Thermal Stress and Force with Restricted Expansion
A DN150 Std. (6 in) steel pipe with length 50 m (1969 in) is heated from 20oC (68oF) to 90oC (194oF). The expansion coefficient for steel is 12 10-6 m/mK (6.7 10-6 in/inoF). The modulus of elasticity for steel is 200 GPa (109 N/m2) (29 106 psi (lb/in2)).
make 3D models with The Engineering ToolBox Sketchup Extension
Expansion of unrestricted pipe:
dl = (12 10-6 m/mK) (50 m) ((90oC) - (20oC))
= 0.042 m
If the expansion of the pipe is restricted - the stress created due to the temperature change can be calculated as
σdt(200 109 N/m2) (12 10-6 m/mK) ((90oC) - (20oC))
= 168 106 N/m2 (Pa)
= 168 MPa
Note! - if there is pressure in the pipe - the axial and circumferential (hoop) stress may be added to restricted temperature expansion stress by using vector addition.
The outside diameter of the pipe is 168.275 mm (6.63 in) and the wall thickness is 7.112 mm (0.28 in). The cross-sectional area of the pipe wall can then be calculated to
A = π ((168.275 mm) / 2)2 - π ((168.275 mm) - 2 (7.112 mm)) / 2)2
= 3598 mm2
= 3.6 10-3 m2
The force acting at the ends of the pipe when it is restricted can be calculated as
F = (168 106 N/m2) (3.6 10-3 m2)
= 604800 N
= 604 kN
##### The calculation in Imperial units
Expansion of unrestricted pipe:
dl = (6.7 10-6 in/inoF) (1669 in) ((194oF) - (68oF))
= 1.4 in
Stress in restricted pipe:
σdt(29 106 lb/in2) (6.7 10-6 in/inoF) ((194oF) - (68oF))
= 24481 lb/in2 (psi)
Cross sectional area:
A = π ((6.63 in) / 2)2 - π ((6.63 in) - 2 (0.28 mm)) / 2)2
= 5.3 in2
Axial force acting at the ends:
F = (24481 lb/in2) (5.3 in2)
= 129749 lb
### Example - Thermal Tensions in Reinforced or Connected Materials
When two materials with different temperature expansion coefficients are connected - as typical with concrete and steel reinforcement, or in district heating pipes with PEH insulation etc. - temperature changes introduces tensions.
This can be illustrated with a PVC plastic bar of 10 m reinforced with a steel rod.
The free expansion of the PVC bar without the reinforcement - with a temperature change of 100 oC - can be calculated from (1) to
dlPVC = (50.4 10-6 m/mK) (10 m) (100 oC)
= 0.054 m
The free expansion of the steel rod with a temperature change of 100 oC - can be calculated from (1) to
dlsteel = (12 10-6 m/mK) (10 m) (100 oC)
= 0.012 m
If we assume that the steel rod is much stronger than the PVC bar (depends on the Young's modulus and the areas of the materials) - the tension in the PVC bar can be calculated from the difference in temperature expansion with (4) as
σPVC = (2.8 109 Pa) (0.054 m - 0.012 m) / (10 m)
= 11.8 106 Pa
= 11.8 MPa
The Tensile Yield Strength of PVC is approximately 55 MPa.
### Thermal Expansion Axial Force Calculator
This calculator can be used to calculate the axial force caused by an object with restricted temperature expansion. The calculator is generic and can be used for both metric and imperial units as long as the use of units are consistent.
## Related Topics
• ### Mechanics
The relationships between forces, acceleration, displacement, vectors, motion, momentum, energy of objects and more.
• ### Statics
Forces acting on bodies at rest under equilibrium conditions - loads, forces and torque, beams and columns.
• ### Temperature Expansion
Thermal expansion and expansion coefficients. Expansion pipes and tubes made of stainless steel, carbon steel, copper, plastics and more.
## Related Documents
• ### Fixed Pipes - Stress vs. Change in Temperature
Temperature changes introduces stress fixed pipes.
• ### Fixed Temperature Points
Temperature points that can be used as a reference for calibration
• ### Linear Thermal Expansion
Online linear temperature expansion calculator.
• ### Metals - Temperature Expansion Coefficients
Thermal expansion coefficients metals.
• ### Metals and Alloys - Young's Modulus of Elasticity
Elastic properties and Young's modulus for metals and alloys like cast iron, carbon steel and more.
• ### Pipes and Tubes - Temperature Expansion
Pipes expands when heated and contracts when cooled and the expansion can be expressed with the expansion equation.
• ### Piping Elbows - Thrust Block Forces
Thrust block forces on pipe bends anchors due to liquid velocities and internal pressures - online resulting force calculator.
• ### Piping Materials - Temperature Expansion Coefficients
Temperature expansion coefficients for materials used in pipes and tubes like aluminum, carbon steel, cast iron, PVC, HDPE and more.
• ### Piping Materials - Temperature Heat Expansion and Cooling Contraction
Expansion and contraction when cast iron, carbon and carbon molybdenum steel, wrought iron, copper, brass and aluminum pipes are heated or cooled.
• ### Pressfit Pipes - Expansion Loops
Temperature expansion loops with pressfit piping.
• ### Shrink-Fits Assembly
Assembly temperatures for shrink-fits.
• ### Solids - Volume Temperature Expansion Coefficients
Cubical expansion coefficients for solids.
• ### Steam Pipes - Thermal Expansion
Thermal expansion of steam pipes heated from room temperature to operation temperature (mm pr. 100 m pipe).
• ### Steel Pipes - Temperature Expansion
Calculate temperature expansion with carbon steel pipes.
• ### Steel Pipes - Thermal Expansion Loops
Calculating and sizing steel pipe thermal expansion loops.
• ### Stress
Stress is force applied on cross-sectional area.
• ### Stress in Thick-Walled Cylinders or Tubes
Radial and tangential stress in thick-walled cylinders or tubes with closed ends - with internal and external pressure.
• ### Stress, Strain and Young's Modulus
Stress is force per unit area - strain is the deformation of a solid due to stress.
• ### Thermal Expansion - Linear Expansion Coefficients
Linear temperature expansion coefficients for common materials like aluminum, copper, glass, iron and many more.
• ### Thermal Expansion of Steam Pipes - (inches)
Expansion of steam pipes heated from room temperature to operation temperature.
• ### Thermoplastics - Physical Properties
Physical properties of thermoplastics like ABS, PVC, CPVC, PE, PEX, PB and PVDF.
• ### Young's Modulus, Tensile Strength and Yield Strength Values for some Materials
Young's Modulus (or Tensile Modulus alt. Modulus of Elasticity) and Ultimate Tensile Strength and Yield Strength for materials like steel, glass, wood and many more.
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## Citation
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Modify the access date according your visit.
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. | 2,220 | 8,919 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2024-30 | latest | en | 0.732957 |
https://study-assistant.com/mathematics/question12840050 | 1,632,719,056,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058263.20/warc/CC-MAIN-20210927030035-20210927060035-00457.warc.gz | 572,356,477 | 59,075 | , 22.06.2019 09:10 flore9584
# This time make a simple coaster that bumps the axis at x=500. remember to make sure that the track rises before it falls
### Another question on Mathematics
Mathematics, 21.06.2019 12:50
Tori examined the pattern of exponents in the table. based on the pattern, which statements are true? check all that apply.
Mathematics, 21.06.2019 14:50
An assembly consists of two mechanical components. suppose that the probabilities that the first and second components meet specifications are 0.87 and 0.84. assume that the components are independent. determine the probability mass function of the number of components in the assembly that meet specifications. x
Mathematics, 21.06.2019 16:00
Enter the number of complex zeros for the polynomial function in the box. f(x) = x+ + 5x² +6
Mathematics, 21.06.2019 17:30
Aplot of land is shaped like a quadrilateral. fences are built on the diagonal to divide the area into 4 sections. what is vi to the nearest tenth? gv = 6.55fv = 5.84vh = 3.27vi = ? | 279 | 1,023 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2021-39 | latest | en | 0.888683 |
http://www.slideserve.com/thaddeus-farrell/interconnection-networks | 1,503,191,055,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105955.66/warc/CC-MAIN-20170819235943-20170820015943-00046.warc.gz | 664,710,531 | 14,510 | 1 / 12
# Interconnection Networks - PowerPoint PPT Presentation
Interconnection Networks. Lecture 5 : January 29 th 2007 Prof. Chung-Kuan Cheng University of California San Diego Transcribed by: Jason Thurkettle. Topics. Graph Construction. Project phase 1: See Z9 IBM Journal Research and Development Jan 2007. Hyper Cube (HC).
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Presentation Transcript
• Lecture 5 : January 29th 2007
Prof. Chung-Kuan Cheng
University of California San Diego
Transcribed by: Jason Thurkettle
• Graph Construction
• Project phase 1: See Z9 IBM Journal Research and Development
Jan 2007
• From A -> B:
Divide Q3 into 2 opposed Q2’s.
Note that there are two paths to any point on a Q2.
Now connect the Q2’s.
• How do you change a hypercube to improve metrics?
• Generalized Hypercube
• Toroidal Mesh Hypercube
• Crossed Hypercube
• Folded Hypercube
• Cube Connected Hypercube
• Q(d1, d2,…,dn) = Kd1 x Kd2 x … Kdn
Note: Kd is the complete graph of the degree derived. Cliques
• 1) d1+d2+…+dn=n Regular
• 2) diameter = Dimension = n
• 3) Connectivity
Toroidal Mesh Hypercube C(d1,d2,…,dn)
• x = x1,x2,…,xn & y = y1,y2,…,yn : are linked
iff
or C(d1,d2,…,dn) = Cd1xCd2x…xCdn
where Cdi is a undirected cycle
• 1) 2n regular
• 2) diameter =
• 3) connectivity = 2n
• 4) # nodes =
• x = x1,x2,…,xn & y = y1,y2,…,yn : are linked
iff
• a) xn…xj+1 = yn…yj+1
• b) xj ≠ yj
• c) xj-1 = yj-1 if j is even
• d) x2i,x2i-1 ~ y2i,y2i-1
e.g. x1x2 ~ y1y2 : {(00,00),(10,10),(01,11),(11,01)}
• 1) 2n vertices, n2n-1 edges
• 2) diameter
• 3) connectivity n
i.e.
• 1) 2n vertices (n+1)2n-1 edges
• 2) n+1 regular
• 3) diameter
• 4) connectivity n+1
• (x,i), (y,j) are linked iff
1) x=y, |i-j|= 1 mod n or
2) i=j, |xi-yi| = 1
Note: 1 & 2 refer to a cycle and not a hypercube.
• 1) n2n vertices – 3n2n-1 edges
• 2) 3 regular
• 3) Diameter=
• Def 1: d-ary sequence of length n
• Def 2: iterated line digraphs
• B(d,1) = Kd+ B(d,n)=Ln-1 (Kd+) n≥2
Note: Kd+ is a complete d vertex graph
• Def 3: V = {0,1,…,dn-1}
E={(x,y),y=xd+β mod dn, β=0,1,…d-1}
• 1) dn vertices, dn+1 edges
• 2) d regular
• 3) diameter = n | 952 | 2,687 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2017-34 | longest | en | 0.744013 |
https://moorejustinmusic.com/students-life/how-much-is-a-yard-of-dirt/ | 1,723,632,224,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641107917.88/warc/CC-MAIN-20240814092848-20240814122848-00495.warc.gz | 304,306,572 | 8,843 | # How much is a yard of dirt?
## How much is a yard of dirt?
Cost to Deliver Landscaping Fill
Material Cost per Cubic Yard Delivery
Topsoil \$12 – \$55 \$15-\$150
Dirt \$5 – \$15 \$150 for 10 – 13 cubic yards
Sand \$15 – \$20 \$50 – \$150
Mulch \$15 – \$65 \$150 for 10-13 cubic yards
## How much does 18 yards of dirt weigh?
Soil: Weighs about 2,200 pounds per cubic yard, depending on the moisture content. Sand, Gravel, Stone: Can tip the scales at upwards of 3,000 pounds per cubic yard.
How much does 18 cubic yards of topsoil cost?
In some areas of the US topsoil sells for about \$12-\$18 a cubic yard, with a delivery charge of \$15-\$60 depending on amount and distance (and delivery may be free with larger loads of 10 yards or more), for a total cost of \$75-\$150 for five yards, delivered.
### How much does a yard of dirt cover?
One cubic yard of soil covers 100 square feet at a 2 inch depth.
### How many pounds of dirt would be in a yard?
In general, a pure cubic yard of fill dirt will weigh somewhere between 2,000 to 2,700 pounds depending on moisture content and composition. When it comes to a pure cubic yard of topsoil dirt, it will weigh around 1,080 pounds.
How do you calculate yard of dirt?
Multiply the three dimensions together to find the number of cubic feet (0.5’ x 12’ x 12 = 72 cubic feet) Divide the cubic feet by the number of cubic feet in a cubic yard (27) to find the number of cubic yards (72 ÷ 27 = 2.67 cu. yd.). Therefore, you would need 2.67 cubic yards of dirt to fill the flower bed. Here’s…
## How many square feet in a yard of dirt?
1 Cubic Yard of dry Dirt is approximately 2,000 lbs. or one ton (this weight will increase significantly when product is wet). 1 Yard of Dirt will cover approximately 324 square feet at a 1 inch depth.
## How much dirt do I need to level my yard?
To determine how much soil you need for your low areas, plan for soil 1-inch deep; this is the most you should raise the level of the dirt in a year’s time, according to the Walter Reeves, Master Gardener website. For a depth of 1 inch, the basic requirement for topsoil is 3 cubic yards, or 81 cubic feet, for every 1,000 square feet of yard. | 585 | 2,189 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2024-33 | latest | en | 0.929725 |
aprllp.com | 1,717,082,581,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971668873.95/warc/CC-MAIN-20240530145337-20240530175337-00257.warc.gz | 76,444,403 | 36,500 | 81 Chancery Lane
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# APR Teasers
On this page we will be adding some quiz questions that we post in our monthly client updates - every time we post a new teaser we will update the previous one with a solution, so look out for these each month.
### July 2023 - A Ghastly take on Schrodinger's Cat
Teaser
A king is having a party in 24 hours.
He has 8 casks of wine in the cellar, but someone has informed him that exactly one of them is poisoned.
The poison will kill whoever drinks it sometime, randomly, within 24 hours.
The king is insistent that he won't waste any wine that he doesn't have to, and he doesn't mind sacrificing some of his servants to make sure that he only has to throw out the one poisoned cask.
What is the fewest number of servants the king will need to find out with certainty which cask is poisoned in time for the party?
### May 2023 - A Technical Knock-out
Teaser
A knockout tournament has 64 players. In each round, the remaining players are paired up randomly. The winners progress to the next round and the losers are eliminated. Assume each player has a different skill level and that better players will always beat worse players. How many different finals are possible?
### March 2023 - Nines in 1 to 1,000,000
Teaser
If you were to write every number from one to one million, how many times would you write the number nine?
### February 2023 - A Game of Dice
Teaser
You roll a fair 6-sided dice twice and take the maximum of the two rolls. What is the expected value of this result? For example, if you rolled a 6 and a 3, your result here would be a 6.
### November 2022 - A Meeting at the Train Station
Teaser
Alice and Bob both go to the railway station between 10 and 11am. Each of them turns up at a uniformly-selected, random time between 10 and 11. They then wait for 10 minutes and leave.
What is the probability that Alice and Bob meet each other at the station? (You may assume they meet if there is any point when they are both at the station).
### September 2022 - Cutting a Straight Line
Teaser
If a straight line is cut independently and uniformly at random in two places, what is the probability that the three segments can form a triangle?
### July 2022 -
Teaser
You drop a circular hoop of diameter d<1 onto an infinite chessboard with squares of side length 1.
What is the probability that the hoop surrounds an area containing just one colour?
### June 2022 -
Teaser
What is the smallest positive integer that is 1 more than a multiple of 2, 2 more than a multiple of 3, 3 more than a multiple of 5 and 4 more than a multiple of 7?
### April 2022 - Jiggly Numbers
Teaser
A Jiggly number is a 4-digit number, no digit of which is zero, such that however the digits of the number are rearranged, the resulting 4-digit number is always a multiple of 12.
How many Jiggly numbers are there?
### March 2022 - Avoiding Awkward Introductions
Teaser
There is a round table with 8 chairs. On each chair there is a person.
How many ways are there that each person can shake their right hand with the right hand of somebody else without two pairs of hands crossing each other?
### February 2022 - A Numerical Poem
Teaser
A dozen, a gross and a score,
Plus three times the square root of four,
Divided by seven,
Plus X times eleven,
Is nine squared and not a bit more.
What is x?
### January 2022 - A Strange Extra-Curricular Activity
Teaser
There is a school with 1,000 students and 1,000 lockers. On the first day of term the head teacher asks the first student to go along and open every locker. He asks the second to go to every second locker and close it. He then asks the third to go to every third locker and close it if it is open or open it if it is closed, and so on. The process is completed with the thousandth student.
How many lockers are open at the end?
### December 2021 - Secret Santa
Teaser
APR’s secret Santa consists of 20 people who each write their name on a piece of paper and put it in a hat. Each person randomly draws a name from the hat to determine who has them as their secret Santa.
What is the probability that at least one person draws their own name?
### November 2021 - The Trouble with Time Travel
Teaser
Channelling your inner Marty McFly, you travel one week back in time in order to win the lottery. The jackpot is worth £1 million, and each ticket costs £1. Note that if you win, your ticket purchase is not refunded.
The problem is, you’re not the only person with time-travelling powers. There are 10 other time travellers who also know the winning numbers. You know for a fact that each of them will buy exactly one lottery ticket. Under the lottery’s rules the jackpot is evenly split among all the winning tickets (i.e. not evenly among winning people).
How many tickets should you buy to maximise your profits, assuming that no non-time-travellers win the lottery?
### October 2021 - APR Mastermind
Teaser
There are 200 episodes in a series of “APR Mastermind”. The first episode of the series features three brand-new contestants. Each subsequent episode includes a returning champion (the winner of the previous episode) as well as two new challengers. Throughout the series, it so happens that the returning champions are particularly strong, with each one winning five consecutive episodes before being dethroned on the sixth.
If you pick a contestant at random from across the series, what is the probability that they are an APR Mastermind champion (meaning they won at least one episode)?
### September 2021 - Exploring a Chess Board
Teaser
A standard 8x8 chessboard has each square marked with a different numeric value according to the following rule: for a given square, its value is the (arithmetic) mean of the other squares on the board from which a knight can reach the given square in one move. (Knights move in a standard "L"-shape.) For example, using standard algebraic chess notation, the value on the square a1 is equal to the mean of the values on squares b3 and c2.
If the square on the top left of the board (square a8) has a value of 21, what is the value of the square on the bottom right of the board (square h1)?
### August 2021 - The Olympics
Teaser
At the Tokyo Olympics a gold medal weighed 556g, a silver medal weighed 550g and a bronze medal weighed 450g.
What was the total weight of all the medals awarded to athletes during the Olympics?
### July 2021 - The Bag of Marbles
Teaser
A bag contains 100 marbles, and each marble is one of three different colours.
If you were to draw three marbles at random, the probability that you would get one of each colour is exactly 20 percent.
How many marbles of each colour are in the bag?
### May 2021 - An Age Old Question
Teaser
Today, Anna is half the age of Chris.
8 years ago, Anna was half the age of Danielle.
9 years ago, Danielle was half the age of Ben.
When Anna was born, Ben was half as old as Anna is now.
How old will each person be when the sum of all of their ages is 65?
### April 2021 - Dominoes and Chess
Teaser
You have been given 31 dominoes, and a chessboard with 62 squares, where two diagonally opposite corners have been cut away. Each half of a domino is the same size as one of the chessboard squares.
Can you find a way to lay the dominoes on the chessboard in such a way that each square is covered?
### March 2021 - Rock-Paper-Scissors: Sudden Death!
Teaser
You're playing rock paper scissors with your friend. You've played together a lot, and you're quite good at guessing what move they will make: you get it right 50% of the time. You're playing a sudden death match: the first to win one round is the winner, and if you tie, you keep playing until someone wins. 'You can assume that an incorrect guess has an equal chance of resulting in either a draw or a loss'.
What is the probability you win?
### February 2021 - The World's Greatest Game-Show
Teaser
Imagine you are a contestant on a game show. The host has randomly generated a number between 0 and 100 (this could be a whole number or a decimal) and asks you to guess a number lower than it. If you successfully guess a lower number, then you win the amount of your guess. If you guess a higher number, then you leave with nothing.
Assuming you want to maximise your expected earnings, what number should you guess?
### January 2021 - A Game of Pool
Teaser
Three students decide to play a round-robin pool tournament, where the winner stays on and the loser waits their turn to play again.
At the end of the evening, they count the number of games that each of them played:
Ash played 9,
Beth played 13,
Craig played 16.
Who lost the second game?
### November 2020 - Group Stage Analysis
Teaser
A Nations cup group consists of 4 teams who all play each other twice. The teams finishing first and second after all the matches have been played qualify for the knockout stage of the competition. 3 points are awarded for a win and 1 for a draw (0 for a loss). In the case of a points tie a scale of other metrics are used to decide final positions such that no team has the same final position in the group.
What is:
a) the lowest number of points a team can get and still qualify for the knockout stage?
b) the highest number of points a team can get and not qualify for the knockout stage?
A. 4 & 11 B. 5 & 11 C. 6 & 11 D. 4 & 12 E. 5 & 12
### October 2020 - Rock-Paper-Scissors
Teaser
Cari and Morgan play rock-paper-scissors 10 times. You know that Cari uses rock three times, scissors six times, and paper once. Morgan uses rock twice, scissors four times, and paper four times. You know there are no ties in all 10 games but the order of games is unknown.
Who wins and by how many games?
A. Morgan by 6 B. Morgan by 3 C. It is a draw D. Cari by 2 E. Cari by 4
### September 2020 - The Largest Slice of Pizza
Teaser
100 people are lined up waiting for a slice of pizza they will all be sharing. The first person takes 1% of the pizza. The second then takes person then takes 2% of what is remaining. The third takes 3% of what is remaining and this continues with the nth person always taking n% of the remaining pizza.
Which number in the line will get the largest slice of pizza?
A. 1 B. 3 C. 5 D. 10 E. 25
### August 2020 - A Pressing Partner Problem
Teaser
The APR partners have decided to have new office name plates printed, in order to further increase the respect in which they are held by APR staff and allow them to spend a few hours imagining that they will get back to their offices this year.
The company providing the name plates charges nothing for the plate itself, and instead charges according to the letters being printed. The following quotes have been obtained:
Roger Austin: £19
Gary Heslop: £15
Tim Nash: £11
How much will the company charge if Chris Bryce decides to purchase a name plate (from his own funds, of course!)?
A. £12 B. £14 C. £15 D. £17 E. £0 (Chris should not have a name plate!)
### July 2020 - A Precarious Bridge Crossing
Teaser
Four people are on one side of a weak bridge at night and need to get across. The bridge will only support the weight of two people at a time and a torch is needed to cross safely, but the group only has a single torch. Each person takes a different amount of time to cross - 10 minutes, 5 minutes, 2 minutes and 1 minute respectively - and any pair crossing the bridge must travel at the speed of the slower person.
What is the shortest time in which all four can get from one side of the bridge to the other?
A. Not possible B. 48 mins C. 21 mins D. 19 mins E. 17 mins
### June 2020 - A Corridor Full of Light Switches
Teaser
100 people walk down a corridor with 100 lights, all initially off. Each person flicks all switches that their number divides. E.g. first person flicks all switches, number 3 flicks 3,6,9,...,99 and the 37th person flicks just 37 and 74.
How many switches are now in the on position after all the people have walked through?
A. 0 B. 10 C. 25 D. 100 E. More information is required
### May 2020 - A Fruit Factory Mix-Up
Teaser
A wholesale fruit factory produces only boxes of apples, boxes of oranges or boxes of mixed apples and oranges. These are labelled by an automated machine however an error has been introduced in the label machines code and now means that every box has been labelled incorrectly.
If you have one of each box which box should you remove a single piece of fruit from to allow you to relabel all boxes correctly.
A. Apples & Oranges B. Apples C. Oranges D. Any of them E. None of them
### April 2020 - How to Break a Chocolate Bar
Teaser
For a bar of chocolate comprising of 30 squares (a large 6 by 5 bar of your favourite brand say) how many times would you need to break the bar to make it into 30 individual squares?
A. 11 B.15 C. 29 D. 30 E. 31
### March 2020 - A Very Cross Problem
Teaser
Here is an ordinary cross. Drawing only 2 straight lines what is the greatest number of pieces you can divide the cross into?
A. 3 B.4 C. 5 D. 6 E. 8
### February 2020 - A Sequence of Tricky Numbers
Teaser
What is the next number in the sequence?
1, 11, 21, 1211, 111221, …
A. 1122221 B. 1212221 C. 312211 D. 1331212 E. 22
### January 2020 - Mathematicians and Their Names
Teaser
The first six volumes of Encyclopaedia of Mathematics are arranged in order on my shelf from left to right. The six volumes contain mathematicians with surnames beginning; A-Ba, Be-Ca, Ce-Ei, Ek-Fe, Fee-Fi and, Fo-Fum respectively. If one ignores the covers, which of the following encyclopaedia entries could be on the page ‘next to’ the page with the entry for Einstein as they sit on my shelf?
A. Abel B. Bernoulli C. Cantor D. Euler E. Fibonacci
### December 2019 - An Excel Conundrum
Teaser
Create the shortest possible formula that can be copied from cell A1 downwards to replicate the game "FizzBuzz".
The game works like this:
• The players count up from 1.
• Each multiple of 3 is replaced with the word "Fizz".
• Each multiple of 5 is replaced with the word “Buzz”.
• If a number is both a multiple of 3 and a multiple of 5, then "FizzBuzz" should be used.
Other rules:
• No VBA allowed.
• Your formula has to be written into the cell - no hiding it in a named range or similar.
• That means no row numbers in a separate column, no inputs in fixed cells, no range names, no variables, no formulae in conditional formatting or text formatting, etc.
• Any Excel version allowed, but no add-in or other non-standard functions.
• Array formulae are fine, but the curly braces add 2 to your character count.
### November 2019 - A Game of Tennis
Teaser
A tennis player has a 60% chance of winning any given point in her service game. Assuming that all points are independent of each other, what is the probability that she will win the service game?
(The scoring system in tennis is such that a player must score a minimum of 4 points and be at least 2 clear points ahead of their opponent in order to win a game.)
### October 2019 - The Prisoners and The Hats
Teaser
One day a jailer brought three prisoners out and had them stand in a line, one behind the other. He then showed them five hats (three red and two white) which he hid in a bag. From these five hats, he selected three and put one on each of the prisoners’ heads. None of the men could see what colour hat he himself wore, but could see the colour of any hats in front of him.
Starting from the back of the line, the jailer asks the first man what colour his hat is, offering him his freedom if he is correct but doubling his sentence if he gets it wrong; he declines to answer. The jailer then asks the second man the colour of his hat, making the same offer; he also declines to answer.
What colour is the third man’s hat? | 4,059 | 16,461 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2024-22 | longest | en | 0.947308 |
http://www.ask.com/question/Convert-83-Kilograms-to-Pounds | 1,394,315,322,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1393999664205/warc/CC-MAIN-20140305060744-00065-ip-10-183-142-35.ec2.internal.warc.gz | 225,699,757 | 18,085 | # Convert 83 Kilograms to Pounds?
There are two pounds and 3.27 ounces in one kilogram, so to convert kilograms to pounds, you will want to make sure it is all in one format first. There are 16 ounces in one pound. So in one kilogram, there are 35.27 ounces in one kilogram. In 83 kilograms, there are 2927.41 ounces. Then to put that back into pounds, you will divide by 16. This will get you 182 pounds and 15.41 ounces.
Reference:
83 kilograms is equal to 182.9837 pounds.
Convert to
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To convert kilograms into pounds, all you need to do is multiply the number of kilograms by 2.20462262185. For instance, 10 kilograms would equal 22.0462262185 pounds. http://answers.ask.com/Reference/Other/how_to_conv...
1. Determine the value in pounds (represented by "lb" in the formula) you want to convert to kilograms and insert it into the blank space below. _ lb. * 1 kg. 2.2046226218 http://www.wikihow.com/Convert-Pounds-to-Kilograms
8 pounds? 83 kg = 182.983 lb. yes. exactly. 182.983678 pounds. http://wiki.answers.com/Q/How_many_pounds_is_83_ki...
1. Convert from pounds to kilograms by taking the number in pounds and dividing it by 2.2. For example, a man weighs 200 lbs. so 200 / 2.2 = approximately 91 kg. 2. Convert from kilograms http://www.ehow.com/how_4578188_convert-pounds-kil...
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There are two pounds and 3.27 ounces in one kilogram, so to convert kilograms to pounds, you will want to make sure it is all in one format first. There are 16 ...
One kilogram is the equivalent to 2.20462262 pounds. You can convert kilograms into pounds by multiplying the number of kilograms by 2.20462262. For instance ...
Pounds can be converted into kilograms by dividing the number of pounds by 2.2046. Pounds and kilograms are used to measure weights like those of a person or a ... | 507 | 1,892 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2014-10 | latest | en | 0.870071 |
https://slideum.com/doc/4071008/no-slide-title | 1,701,849,393,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100583.31/warc/CC-MAIN-20231206063543-20231206093543-00219.warc.gz | 600,027,410 | 8,154 | #### Transcript No Slide Title
```Unit 27
Transformers
• Objectives
– After completing this chapter, the student
should be able to:
•
•
•
•
Describe how a transformer operates.
Explain how transformers are rated.
Explain how transformers operate in a circuit.
Describe the differences between step-up, stepdown, and isolation transformers.
• Describe how the ratio of the voltage, current,
and number of turns are related with a
transformer.
• Describe applications of a transformer.
• Identify different types of transformers.
• Electromagnetic induction
– The action caused when two electrically
isolated coils are placed next to each other and
an AC voltage is put across one coil, resulting
in a changing magnetic field which induces a
voltage into the second coil..
– The device used to create this action is called a
transformer.
• Transformers
– The coil containing the AC voltage is the
primary winding.
– The coil in which the voltage is induced is the
secondary winding.
– Coefficient of coupling
• A number from 0 to 1.
– 1 indicating that all the primary flux lines cut the
secondary windings.
– 0 indicating that none of the primary flux lines cut the
windings.
• The design of a transformer is determined
by:
– The frequency at which it will be used.
• Low-frequency applications use iron cores.
• High-frequency applications use air cores.
– The power it must handle.
– The voltage it must handle.
• Transformers are rated in volt-amperes.
• Transformers are wound with tapped
secondaries.
– Center tapped secondary is equal to two
secondary windings.
– Used for power supply to convert AC voltages
to DC voltages.
• Mutual inductance
– The primary induces a voltage into the
secondary and the secondary induces a voltage
back into the primary.
• Turns ratio
– Determines whether a transformer is used to
step up, step down, or pass voltage unchanged.
– The number turns in the secondary winding
divided by the number of turns in the primary
winding.
NS
– Expressed as: turns ratio =
NP
– where N = number of turns.
• A step-up transformer
– A transformer with secondary voltage greater
than its primary voltage.
– Expressed as:
ES NS
EP NP
– The turns ratio is always greater than one.
• A step-up transformer
– A transformer that produces a secondary
voltage less than its primary voltage.
– The turns ratio is always less than one.
• When a transformer steps up the voltage, it
steps down the current.
– This is expressed as:
PP = P S
(IP)(EP) = (IS)(ES)
• The current is inversely proportional to the
turns ratio.
This is expressed as:
IP NS
IS N P
• Impedance ratio is equal to the turns ratio
squared.
– This is expressed as:
2
ZP N P
2
ZS N S
• Applications for transformers
–
–
–
–
–
–
–
Stepping up voltage and current.
Stepping down voltage and current.
Impedance matching.
Phase shifting.
Isolation.
Blocking DC while passing AC.
Producing several signals at various voltage
levels.
• Transformers are used for:
– Transmitting electrical power to homes and
industry.
– Isolating electronic equipment from 120-volts
AC, 60-hertz power while it is being tested.
• Does not step up or step down the voltage.
• Autotransformers
– A device used to step up or step down applied
voltage.
– Both the primary and secondary windings are
part of the same core.
• In Summary
– Transformers consist of:
• two coils
• a primary winding
• a secondary winding
– Transformers allow an AC signal to be
transferred from one circuit to another.
– Transformers allow:
• stepping up the signal.
• stepping down the signal.
• passing the signal unchanged.
– Transformers are designed to operate at certain
frequencies.
– Transformers are rated in volt-amperes.
– Turns ratio determines whether a transformer is
used to:
• step up a voltage.
• step down voltage.
• pass voltage unchanged.
– A step-up transformer:
• produces a secondary voltage greater than its
primary voltage.
• has a turns ratio that is always greater than one.
– A step-down transformer:
• produces a secondary voltage less than its
primary voltage.
• has a turns ratio that is always less than one.
– The turns ratio determines the amount of
voltage that is stepped up or down.
– Transformer applications include:
• Impedance matching.
• Phase shifting.
• Isolation.
• Blocking DC while passing AC.
• Producing several signals at different voltage
levels.
– Isolation transformers
• Pass the signal unchanged.
• Used to prevent electric shocks.
– Autotransformers
• Used to step up or step down voltage.
• Do not provide isolation.
``` | 1,058 | 4,516 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2023-50 | latest | en | 0.906925 |
http://essay100.fam.cx/page/two_tailed_hypothesis_testing_examples/ | 1,579,424,735,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250594333.5/warc/CC-MAIN-20200119064802-20200119092802-00166.warc.gz | 62,963,329 | 4,679 | two tailed hypothesis testing examples
One tailed or two tailed Hypothesis? A one-tailed directional hypothesis predicts the nature of the effect of the independent variable on the dependent variable in statistical significance testing, a one-tailed test and a two-tailed test are alternative ways of computing the statistical significance of a parameter inferred. An introduction to one and two tail tests used in hypothesis testing using a standard bell curve with a population mean and sample mean psychology definition for two tailed hypothesis in normal everyday language, edited by psychologists, professors and leading students. It includes help us get better. One- and two-tailed tests A two-tailed test applied to the an introductory statistics text for the social sciences. He explains this as because a specific set of data may be unlikely (in the null hypothesis) introductory statistics: concepts, models, and applications lang t, altman d. Sal continues his discussion of the effect of a drug to one-tailed and two-tailed hypothesis tests statistical analyses and methods in the published literature: the sampl guidelines. Summary 2 comprehensive and comprehensible set of there’s a lot of controversy around one-tailed vs two-tailed testing. One of the main goals of statistical hypothesis testing is to estimate the P value, which is the probability of obtaining the observed results, or something in fact, many people don’t realize that there are two ways to determine whether an. One-Tailed and Two-Tailed Tests Watch the next lesson: actually, whenever i talk about an hypothesis, i am really thinking simultaneously about two hypotheses. One- and Two-Tailed Tests we say you have a two-tailed hypothesis. Author(s) David M one and two tailed tests. Lane see also: hypothesis testing suppose we have a null hypothesis h0 and an alternative hypothesis h1. Prerequisites we consider the distribution given by. Binomial Distribution, Introduction to Hypothesis Testing, Statistical one- and two-tailed tests. Describes how to test the null hypothesis that some estimate is due to chance vs the alternative hypothesis that there is some statistically significant effect the test of such a hypothesis is nondirectional or two‐tailed because an extreme test statistic in either tail of the distribution. Sample Size p values and confidence intervals speaking of confidence intervals, let s bring them back into the picture. • • Normal Approximation What is hypothesis testing?(cont it s possible to show that the two definitions of. ) The hypothesis we want to test is if H 1 is \likely true what does statistical significance really mean? many researchers get very excited when they have discovered a statistically significant. So, there are two possible outcomes: Reject H 0 and accept 1 because of a statistical test in which the critical area of a distribution is two sided and tests. How to conduct a hypothesis test to determine whether the difference between two mean scores is significant two-tailed test a two-tailed.
One- and two-tailed tests - Wikipedia
One-Tailed Vs. Two-Tailed Tests: Differences & Examples.
Two-Tailed Test - Investopedia
two tailed hypothesis testing examples
Rating 5 stars - 697 reviews
Sal continues his discussion of the effect of a drug to one-tailed and two-tailed hypothesis tests statistical analyses and methods in the published literature: the sampl guidelines.
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http://math.stackexchange.com/questions/tagged/closed-form+recurrence-relations | 1,408,722,742,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500824209.82/warc/CC-MAIN-20140820021344-00397-ip-10-180-136-8.ec2.internal.warc.gz | 132,634,355 | 22,459 | # Tagged Questions
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### Topic: if statement evaluating complex expressions (Read 963 times)previous topic - next topic
#### yerg2k
##### Jan 22, 2008, 03:36 amLast Edit: Jan 22, 2008, 04:39 am by yerg2k Reason: 1
Hi fellow coders, I'm having trouble wrapping my mind around a confusing issue and need some clarification regarding how if statements are evaluating.
The following code works, but it shouldn't:
Code: [Select]
`if ((preamble == 150) && ((curdata >> 16) == ~(curdata & 0xFFFF))) { return(curdata & 0xFFFF); } `
In my transmitter code, the integer to be transmitted is shifted out, followed by its inverse. Here is the original thread in the hardware forum. In that thread,
Quote
I really don't know how it works like this. If curdata == 0xAAAA5555 and you put it through
Code: [Select]
`((curdata >> 16) == ~(curdata & 0xFFFF))`
then (curdata >> 16) = 0x0000AAAA
and ~(curdata & 0xFFFF) = 0xFFFFAAAA
this is because curdata & 0xFFFF evaluates first giving 0x00005555, then the ~ is evaluated giving 0xFFFFAAAA.
I don't understand how that expression can ever evaluate to true. If somebody knows that part of the puzzle please let me know! [smiley=huh.gif]
you might want to try something like this:
Code: [Select]
`(((curdata >> 16) & 0xFFFF) == (~curdata & 0xFFFF))`
Again this code shouldn't work, but does. It is returning (curdata & 0xFFFF) which should be the real integer's inverse. But, it is returning the actual number sent from the transmitter.
The following code still works like above, but is properly put together and reads correctly:
Code: [Select]
` if ((preamble == 150) && ((curdata >> 16) == (~curdata & 0xFFFF))) { return(curdata >> 16); } `
The two code snippets work the same, except the former returns (curdata & 0xFFFF) which really should be the inverse of the actual number, and the latter code returns it properly (curdata >> 16). Something about the way the if statement evaluates, is causing it to work when it shouldn't. It has something to do with ~(expression) changing the value held in curdata methinks.
Could someone please explain why the code above works like it does? Thanks In Advance.
#### mem
#1
##### Jan 22, 2008, 08:10 amLast Edit: Jan 22, 2008, 08:11 am by mem Reason: 1
The code as written should and does work because curdata is really two consecutive integers treated as a long, where the second integer is equal to the first integer with the bits inverted
(curdata >> 16) == ~(curdata & 0xFFFF)
means that the non inverted int value ie :
(curdata >> 16)
is the same as the second integer inverted
(curdata & 0xFFFF) is the second integer and the tilde ~ inverts this, which undoes the inversion performed in the send.
I hope that makes sense.
#### yerg2k
#2
##### Jan 22, 2008, 04:03 pm
So, then I guess both
(curdata >> 16)
and
~(curdata & 0xFFFF)
result in unsigned ints that get compared? If they resulted in unsigned long then the first statement would evaluate as 0000nnnn and the second as FFFFnnnn, since the leading 0's would be inverted. This is obviously not the case.
Regardless, it is working right in the "fixed" version with (~curdata & 0xFFFF) which is good.
#### mem
#3
##### Jan 22, 2008, 04:31 pm
If you are interested in a peek at the assembler code the compiler produces here it is:
Code: [Select]
`if ( ((curdata >> 16) == ~(curdata & 0xFFFF))) { e2: 29 81 ldd r18, Y+1 ; 0x01 //load least significant byte of curdata e4: 3a 81 ldd r19, Y+2 ; 0x02 // into working registers e6: 4b 81 ldd r20, Y+3 ; 0x03 e8: 5c 81 ldd r21, Y+4 ; 0x04 // this is most significant byte ea: 89 81 ldd r24, Y+1 ; 0x01 // this loads a copy of curdata ec: 9a 81 ldd r25, Y+2 ; 0x02 //into another set of registers ee: ab 81 ldd r26, Y+3 ; 0x03 f0: bc 81 ldd r27, Y+4 ; 0x04 f2: 9a 01 movw r18, r20 // move the upper two bytes to the lower (i.e. >>16) f4: 44 27 eor r20, r20 // exclusive or upper bytes f6: 55 27 eor r21, r21 f8: a0 70 andi r26, 0x00 ; 0 // set the copy of upper bytes to 0 fa: b0 70 andi r27, 0x00 ; 0 fc: 80 95 com r24 // flip all the bits in the copy fe: 90 95 com r25 100: a0 95 com r26 102: b0 95 com r27 104: 28 17 cp r18, r24 // compare the four bytes 106: 39 07 cpc r19, r25 108: 4a 07 cpc r20, r26 10a: 5b 07 cpc r21, r27 10c: 41 f4 brne .+16 ; 0x11e <loop+0x60> //branch if not equal`
#### CasNet
#4
##### Jan 26, 2008, 12:25 am
If curdata is defined as a signed integer then the the ">>" operator may be doing sign extension so that the MSB in 0xAAAA5555 is being shifted right to be 0xFFFFAAAA.
Regards,
David
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# 三角形 in English:
## 1. triangle
The sum of all the angles in a triangle equals 180 degrees.
It must be a triangle.
Having sex in triangle means two woman and one man
The teacher drew a triangle on the blackboard.
A triangle has three sides and three angles.
In the Bermuda Triangle there is a parallel universe.
If a triangle has two right angles, it's a square missing one side.
God cannot make a triangle with more than 180 degrees.
Please find the area of the triangle.
The child drew a spherical triangle.
For example, Pepperberg would show Alex an object, such as a green wooden peg or a red paper triangle.
Pythagoras' theorem allows you to calculate the length of the hypotenuse of a right triangle.
Christopher Columbus was once quoted as saying that pirates were too "simple-minded". He created the Bermuda Triangle later that year.
#### English word "三角形"(triangle) occurs in sets:
Shapes names in Chinese | 204 | 923 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2020-24 | latest | en | 0.930499 |
https://automl.github.io/fanova/fanova_starter.html | 1,722,932,646,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640476915.25/warc/CC-MAIN-20240806064139-20240806094139-00028.warc.gz | 81,661,663 | 3,842 | fANOVA for Starters¶
Introduction¶
Whereas many algorithms presumably depend on a large hyperparameter space, it is known that in many cases only few parameter changes can be responsible for nearly all of the performance improvement. To this end, different techniques have been proposed that score parameters based on their importance. The functional analysis of variance (fANOVA) uses an empirical performance model (EPM) which is based on random forests in order to analyze how much of the performance variance in the configuration space is explained by single parameters or combinations of few parameters.
Note that fANOVA’s result is ideally used for minimizing the hyperparameter search space and should not be seen as a flawless parameter configurator.
More detailed information in : “An Efficient Approach for Assessing Hyperparameter Importance” by Frank Hutter, Holger Hoos and Kevin Leyton-Brown
In order to interpret your plots, you should have in mind what kind of evaluation measurement function you wanted to consider: Either you’d like to maximize or minimize your function. Therefore, depending on this, you’d have to consider low or high performance values.
Here we have the perplexity as measurement. Thus we would like to reduce it.
First, by looking at the importance of each parameter we can clearly see that parameter Col2 is marginally most important. So the result states that the Col2 parameter by itself is responsible for approx. 62% of the perplexity’s variability across the entire space. On the contrary, Col1 is marginally less important with 3,5%.
Parameter Importance
Col0 0.066
Col1 0.035
Col2 0.619
And by looking at the pairwise marginals we have the combination of Col0 with Col2 as marginally most important.
Parameter pair Importance
[Col0, Col1] 0.011
[Col0, Col2] 0.146
[Col1, Col2] 0.067
Now let’s take a look at the visualizations thereof:
Since Col2 was chosen as marginally more important, we will take a closer look at its single marginal plot. It shows that large values for the parameter consistently yield lower performance (in our case perplexity).
In order to capture interaction effects we take a closer look at the pairwise marginal plots of Col0 and Col2.
Here we can see that Col0 is much more important for smaller values of Col2 than for larger ones. Such an interaction cannot be shown by single marginals.
Note that nevertheless this example consists of a lower-dimensional dataset (3 parameters), fANOVA can still give interesting insights. But you should have in mind that it is important to have a large dataset consisting of enough examples so that fANOVA can deliver significant and interpretable results.
Manual
Citing fANOVA | 575 | 2,707 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-33 | latest | en | 0.871933 |
https://www.racedepartment.com/threads/g25-settings.435/ | 1,600,677,723,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400201601.26/warc/CC-MAIN-20200921081428-20200921111428-00073.warc.gz | 1,031,030,758 | 24,952 | # G25 Settings
#### Steve Jones
##### 75RPM
Hi
I would really like to know what your settings are in Windows and ingame for the G25. I'm struggling to find a configuration i'm comfortable with.
Any help would be appreciated.
Cheers
#### Attila Domján
##### 9000RPM
Well, maybe not the answer what you are looking for, but when i had it my setup was:
At Windows:
- turn off FF
- set up centering spring to 135% strength
- wheel roataion to 180 degrees
Ingame:
- turn off FF
That was the best for me. Yeah, i know, i killed all those nice force feedback effects....
#### Steve Jones
##### 75RPM
Thats interesting m8
I used to race Nascar 2003 without FF because i could never really understand what it was trying to tell me. Over the years playing GTR etc i've adjusted to the FF but i'm still not happy.
I like the idea of putting the centre spring up - at least that way you dont have a loose wheel.
I will try it and see how i get on.
#### Jamie Wilson
##### 500RPM
The FF output from the game should, if it's done properly, act as a centering spring - If you want to take full advantage of the FF system in a given game, you should have it set to 0%. You can have it on to suppliment the FF, but I'm personally against the idea as I prefer the idea of getting exactly what the game is telling me.
Spring effect strength and damper effect strength should be set to 0% - they're more there for compatability with old games than anything else. Your global effects strength though should be set to 101% due to a tiny glitch in the drivers; you'll get slightly more nuanced FF that way.
As for steering wheel rotation, I've done a bit of research on this recently and come up with a few things:
(2 x [Max] Steering Lock) x Steer Ratio = Rotation
Rotation/(2 x Steer Ratio) = Steering Lock
Ratio = Rotation/(2 x Steering Lock)
If you want to emulate the car you're driving 1:1, you need to know it's steering ratio. A road going BMW E46 has a ratio of 15.4:1 (3 turns), I happen to know, so:
1080/(2 x 19.6) = 35.0 ideal lock, if we had a 1080 degree wheel.
'Wrapping' this to a maximum allowed lock (thanks Simbin! :/) of 23, the ideal rotation would be:
2 x 23 x 15.4 = 708.4 ideal rotation.
If you just want roughly the right feel for GT/Touring cars, then start off with 540/23 and if that's too snappy, go to 640-680 or so. If the rotation of the in-game wheel is set, to, say, 270 degrees, you can use the equations above to work out the ideal steering lock value for that rotation.
Hope that helped.
#### Steve Jones
##### 75RPM
Thanks Jamie, thats very helpful m8.
I know its all down to personal preference but it nice to know what other guys are doing.
I'll see how i get on.
#### Mark Reynolds
##### Physics & AI Programmer
If you want to emulate the car you're driving 1:1, you need to know it's steering ratio. A road going BMW E46 has a ratio of 15.4:1 (3 turns),
Just interested, where did you get that from mate ?
#### Jamie Wilson
##### 500RPM
A lot of googling - that's the hardest thing to find out.
It's a case of having to settle for roughly the right feel for WTCC/F3000/FBMW cars, by observing onboard footage and noting when they hit the lockstops and scaling that down accordingly to match the 23 degree limit.
I have, though, managed to alter the rotation of the in-game wheel, but only in the offline version where the files (*.inccar) are accessible. It might work with the online one but it could cause mismatches. It's all a bit smoke-and-mirrors, it'd be much easier if they just gave us a damn option.
If you're wondering what rotation the in-game wheel uses by default btw, WTCC cars use 242 and the F3K/FBMW/Radicals use 195. Interestingly both the Radicals and the Caterhams' steering lock value goes way up to 38 with that tiny wheel rotation which looks just a little bit more than silly on the road.
#### Mark Reynolds
##### Physics & AI Programmer
Ok , well I have an E46 sitting outside thats all
My own guestimates are as following.
From what I can see with a crude protractor, I have around 35 degrees angle on the road wheels and around 990 degrees of lock.
That in theory should roughly equate to 900/32 and 450/16 and 270/10 etc etc
C
#### charliebrown
Sorry guys, I hate to bring this back from the dead but how do you guys change between 900 degrees and 540 degrees of setting in between changing cars? Is there any other way other than quitting the game and going into the logitech profiler to make the changes and then re-launching the game again?
I'm usually online so it's a different car every race, would suck to quit the game everytime i wanted to make a change.
#### Simon Trendell
##### 5000RPM
Unfortunately not, you can Alt-Tab and change the rotation which saves restarting the game.
#### Bob Clayton
##### 1RPM
I know its an old thread, someone may like this info tho
I wanted to have seperate startups for Race 07 so I could set set the wingman profiler to different degrees for different cars...
So I copied the Race07.EXE file to the desktop,
I renamed it Formula07.exe then moved it back into the Race 07 Directory, I then did did the same again naming it RadCat07.exe and again Naming it OldSchool.exe
I then went into the Logitech Profiler created "new profile" and selected the Formula07.exe and gave it a rotation of 420 deg
RadCat07.exe and gave it a rotation of 792 deg
Oldschool.exe and gave it a rotation of 900 deg
I then made a folder and put shortcuts to each exe file, I now open that folder and decide which type of car I am going to drive and use that shortcut to start the game..
eg If I want to drive formula3000 I select Formula07.exe which launches the game with 420 deg steering or if I want to drive the caterham i use the RadCat.exe to launch the game and I have 792 deg rotation
I also had to go into nvidia control panel and set up graphics profile the way I like it for each exe file | 1,475 | 5,923 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2020-40 | latest | en | 0.954291 |
https://dipy.org/documentation/1.6.0./reference/dipy.core/ | 1,679,821,877,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945440.67/warc/CC-MAIN-20230326075911-20230326105911-00512.warc.gz | 247,876,973 | 75,189 | # core
Core objects
test([label, verbose, extra_argv, doctests, ...]) Run tests for module using nose.
## Module: core.benchmarks.bench_sphere
Benchmarks for sphere
Run all benchmarks with:
import dipy.core as dipycore
dipycore.bench()
With Pytest, Run this benchmark with:
pytest -svv -c bench.ini /path/to/bench_sphere.py
Methods func_minimize_adhoc(init_hemisphere, ...) func_minimize_scipy(init_pointset, ...)
## Module: core.geometry
Utility functions for algebra etc
cart2sphere(x, y, z) Return angles for Cartesian 3D coordinates x, y, and z cart_distance(pts1, pts2) Cartesian distance between pts1 and pts2 circumradius(a, b, c) a, b and c are 3-dimensional vectors which are the vertices of a triangle. compose_matrix([scale, shear, angles, ...]) Return 4x4 transformation matrix from sequence of transformations. Compose multiple 4x4 affine transformations in one 4x4 matrix decompose_matrix(matrix) Return sequence of transformations from transformation matrix. dist_to_corner(affine) Calculate the maximal distance from the center to a corner of a voxel, given an affine euler_matrix(ai, aj, ak[, axes]) Return homogeneous rotation matrix from Euler angles and axis sequence. Test whether all points on a unit sphere lie in the same hemisphere. Lambert Equal Area Projection from cartesian vector to plane lambert_equal_area_projection_polar(theta, phi) Lambert Equal Area Projection from polar sphere to plane Least squares positive semi-definite tensor estimation normalized_vector(vec[, axis]) Return vector divided by its Euclidean (L2) norm perpendicular_directions(v[, num, half]) Computes n evenly spaced perpendicular directions relative to a given vector v rodrigues_axis_rotation(r, theta) Rodrigues formula sph2latlon(theta, phi) Convert spherical coordinates to latitude and longitude. sphere2cart(r, theta, phi) Spherical to Cartesian coordinates sphere_distance(pts1, pts2[, radius, ...]) Distance across sphere surface between pts1 and pts2 vec2vec_rotmat(u, v) rotation matrix from 2 unit vectors vector_cosine(vecs1, vecs2) Cosine of angle between two (sets of) vectors vector_norm(vec[, axis, keepdims]) Return vector Euclidean (L2) norm
## Module: core.gradients
GradientTable(gradients[, big_delta, ...]) Diffusion gradient information HemiSphere([x, y, z, theta, phi, xyz, ...]) Points on the unit sphere. auto_attr(func) Decorator to create OneTimeProperty attributes. btens_to_params(btens[, ztol]) Compute trace, anisotropy and assymetry parameters from b-tensors. check_multi_b(gtab, n_bvals[, non_zero, bmag]) Check if you have enough different b-values in your gradient table deprecate_with_version(message[, since, ...]) Return decorator function function for deprecation warning / error. disperse_charges(hemi, iters[, const]) Models electrostatic repulsion on the unit sphere generate_bvecs(N[, iters]) Generates N bvectors. get_bval_indices(bvals, bval[, tol]) Get indices where the b-value is bval gradient_table(bvals[, bvecs, big_delta, ...]) A general function for creating diffusion MR gradients. gradient_table_from_bvals_bvecs(bvals, bvecs) Creates a GradientTable from a bvals array and a bvecs array A general function for creating diffusion MR gradients. gradient_table_from_qvals_bvecs(qvals, ...) A general function for creating diffusion MR gradients. inv(a[, overwrite_a, check_finite]) Compute the inverse of a matrix. orientation_from_string(string_ornt) Return an array representation of an ornt string. Return a string representation of a 3d ornt. ornt_mapping(ornt1, ornt2) Calculate the mapping needing to get from orn1 to orn2. params_to_btens(bval, bdelta, b_eta) Compute b-tensor from trace, anisotropy and assymetry parameters. polar(a[, side]) Compute the polar decomposition. reorient_bvecs(gtab, affines[, atol]) Reorient the directions in a GradientTable. reorient_on_axis(bvecs, current_ornt, new_ornt) reorient_vectors(bvecs, current_ornt, new_ornt) Change the orientation of gradients or other vectors. round_bvals(bvals[, bmag]) "This function rounds the b-values unique_bvals(bvals[, bmag, rbvals]) This function gives the unique rounded b-values of the data unique_bvals_magnitude(bvals[, bmag, rbvals]) This function gives the unique rounded b-values of the data unique_bvals_tolerance(bvals[, tol]) Gives the unique b-values of the data, within a tolerance gap vec2vec_rotmat(u, v) rotation matrix from 2 unit vectors vector_norm(vec[, axis, keepdims]) Return vector Euclidean (L2) norm warn(/, message[, category, stacklevel, source]) Issue a warning, or maybe ignore it or raise an exception.
## Module: core.graph
A simple graph class
A simple graph class
## Module: core.histeq
histeq(arr[, num_bins]) Performs an histogram equalization on arr.
## Module: core.interpolation
Interpolator(data, voxel_size) Class to be subclassed by different interpolator types NearestNeighborInterpolator(data, voxel_size) Interpolates data using nearest neighbor interpolation OutsideImage TriLinearInterpolator(data, voxel_size) Interpolates data using trilinear interpolation interp_rbf Interpolate data on the sphere, using radial basis functions. interpolate_scalar_2d(image, locations) Bilinear interpolation of a 2D scalar image interpolate_scalar_3d(image, locations) Trilinear interpolation of a 3D scalar image interpolate_scalar_nn_2d(image, locations) Nearest neighbor interpolation of a 2D scalar image interpolate_scalar_nn_3d(image, locations) Nearest neighbor interpolation of a 3D scalar image interpolate_vector_2d(field, locations) Bilinear interpolation of a 2D vector field interpolate_vector_3d(field, locations) Trilinear interpolation of a 3D vector field map_coordinates_trilinear_iso Trilinear interpolation (isotropic voxel size) nearestneighbor_interpolate trilinear_interp(data, index, voxel_size) Interpolates vector from 4D data at 3D point given by index trilinear_interpolate4d Tri-linear interpolation along the last dimension of a 4d array
## Module: core.ndindex
as_strided(x[, shape, strides, subok, writeable]) Create a view into the array with the given shape and strides. ndindex(shape) An N-dimensional iterator object to index arrays.
## Module: core.onetime
Descriptor support for NIPY.
Redistribution and use in source and binary forms, with or without modification, are permitted provided that the following conditions are met:
• Redistributions of source code must retain the above copyright
notice, this list of conditions and the following disclaimer.
• Redistributions in binary form must reproduce the above
copyright notice, this list of conditions and the following disclaimer in the documentation and/or other materials provided with the distribution.
• Neither the name of the NIPY Developers nor the names of any
contributors may be used to endorse or promote products derived from this software without specific prior written permission.
THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS “AS IS” AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE ARE DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT OWNER OR CONTRIBUTORS BE LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE.
Utilities to support special Python descriptors [1,2], in particular the use of a useful pattern for properties we call ‘one time properties’. These are object attributes which are declared as properties, but become regular attributes once they’ve been read the first time. They can thus be evaluated later in the object’s life cycle, but once evaluated they become normal, static attributes with no function call overhead on access or any other constraints.
A special ResetMixin class is provided to add a .reset() method to users who may want to have their objects capable of resetting these computed properties to their ‘untriggered’ state.
### References
[2] Python data model, http://docs.python.org/reference/datamodel.html
A descriptor to make special properties that become normal attributes. A Mixin class to add a .reset() method to users of OneTimeProperty. auto_attr(func) Decorator to create OneTimeProperty attributes.
## Module: core.optimize
A unified interface for performing and debugging optimization problems.
NonNegativeLeastSquares(*args, **kwargs) A sklearn-like interface to scipy.optimize.nnls Optimizer(fun, x0[, args, method, jac, ...]) Attributes: PositiveDefiniteLeastSquares(m[, A, L]) Methods SKLearnLinearSolver(*args, **kwargs) Provide a sklearn-like uniform interface to algorithms that solve problems of the form: $$y = Ax$$ for $$x$$ Version(version) This class abstracts handling of a project's versions. minimize(fun, x0[, args, method, jac, hess, ...]) Minimization of scalar function of one or more variables. optional_package(name[, trip_msg]) Return package-like thing and module setup for package name sparse_nnls(y, X[, momentum, step_size, ...]) Solve y=Xh for h, using gradient descent, with X a sparse matrix. spdot(A, B) The same as np.dot(A, B), except it works even if A or B or both are sparse matrices.
## Module: core.profile
Class for profiling cython code
Profiler([call]) Profile python/cython files or functions optional_package(name[, trip_msg]) Return package-like thing and module setup for package name
## Module: core.rng
Random number generation utilities.
LEcuyer([s1, s2]) Return a LEcuyer random number generator. WichmannHill1982([ix, iy, iz]) Algorithm AS 183 Appl. WichmannHill2006([ix, iy, iz, it]) Wichmann Hill (2006) random number generator. architecture([executable, bits, linkage]) Queries the given executable (defaults to the Python interpreter binary) for various architecture information. floor(x, /) Return the floor of x as an Integral.
## Module: core.sphere
HemiSphere([x, y, z, theta, phi, xyz, ...]) Points on the unit sphere. Sphere([x, y, z, theta, phi, xyz, faces, edges]) Points on the unit sphere. auto_attr(func) Decorator to create OneTimeProperty attributes. cart2sphere(x, y, z) Return angles for Cartesian 3D coordinates x, y, and z disperse_charges(hemi, iters[, const]) Models electrostatic repulsion on the unit sphere disperse_charges_alt(init_pointset, iters[, tol]) Reimplementation of disperse_charges making use of scipy.optimize.fmin_slsqp. euler_characteristic_check(sphere[, chi]) Checks the euler characteristic of a sphere faces_from_sphere_vertices(vertices) Triangulate a set of vertices on the sphere. remove_similar_vertices(vertices, theta[, ...]) Remove vertices that are less than theta degrees from any other sphere2cart(r, theta, phi) Spherical to Cartesian coordinates unique_edges(faces[, return_mapping]) Extract all unique edges from given triangular faces. unique_sets(sets[, return_inverse]) Remove duplicate sets. vector_norm(vec[, axis, keepdims]) Return vector Euclidean (L2) norm
## Module: core.sphere_stats
Statistics on spheres
permutations(iterable[, r]) Return successive r-length permutations of elements in the iterable. Computes the cosine distance of the best match between points of two sets of vectors S and T Computes the mean cosine distance of the best match between points of two sets of vectors S and T (angular similarity) eigenstats(points[, alpha]) Principal direction and confidence ellipse random_uniform_on_sphere([n, coords]) Random unit vectors from a uniform distribution on the sphere.
## Module: core.subdivide_octahedron
Create a unit sphere by subdividing all triangles of an octahedron recursively.
The unit sphere has a radius of 1, which also means that all points in this sphere (assumed to have centre at [0, 0, 0]) have an absolute value (modulus) of 1. Another feature of the unit sphere is that the unit normals of this sphere are exactly the same as the vertices.
This recursive method will avoid the common problem of the polar singularity, produced by 2d (lon-lat) parameterization methods.
HemiSphere([x, y, z, theta, phi, xyz, ...]) Points on the unit sphere. create_unit_hemisphere([recursion_level]) Creates a unit sphere by subdividing a unit octahedron, returns half the sphere. create_unit_sphere([recursion_level]) Creates a unit sphere by subdividing a unit octahedron.
## Module: core.wavelet
afb3D(x, af1[, af2, af3]) 3D Analysis Filter Bank afb3D_A(x, af, d) 3D Analysis Filter Bank cshift3D(x, m, d) 3D Circular Shift dwt3D(x, J, af) 3-D Discrete Wavelet Transform idwt3D(w, J, sf) Inverse 3-D Discrete Wavelet Transform Function generating inverse of the permutation sfb3D(lo, hi, sf1[, sf2, sf3]) 3D Synthesis Filter Bank sfb3D_A(lo, hi, sf, d) 3D Synthesis Filter Bank
### test
dipy.core.test(label='fast', verbose=1, extra_argv=None, doctests=False, coverage=False, raise_warnings=None, timer=False)
Run tests for module using nose.
Parameters:
label{‘fast’, ‘full’, ‘’, attribute identifier}, optional
Identifies the tests to run. This can be a string to pass to the nosetests executable with the ‘-A’ option, or one of several special values. Special values are:
• ‘fast’ - the default - which corresponds to the nosetests -A option of ‘not slow’.
• ‘full’ - fast (as above) and slow tests as in the ‘no -A’ option to nosetests - this is the same as ‘’.
• None or ‘’ - run all tests.
• attribute_identifier - string passed directly to nosetests as ‘-A’.
verboseint, optional
Verbosity value for test outputs, in the range 1-10. Default is 1.
extra_argvlist, optional
List with any extra arguments to pass to nosetests.
doctestsbool, optional
If True, run doctests in module. Default is False.
coveragebool, optional
If True, report coverage of NumPy code. Default is False. (This requires the coverage module).
raise_warningsNone, str or sequence of warnings, optional
This specifies which warnings to configure as ‘raise’ instead of being shown once during the test execution. Valid strings are:
• “develop” : equals (Warning,)
• “release” : equals (), do not raise on any warnings.
timerbool or int, optional
Timing of individual tests with nose-timer (which needs to be installed). If True, time tests and report on all of them. If an integer (say N), report timing results for N slowest tests.
Returns:
resultobject
Returns the result of running the tests as a nose.result.TextTestResult object.
Notes
Each NumPy module exposes test in its namespace to run all tests for it. For example, to run all tests for numpy.lib:
>>> np.lib.test()
Examples
>>> result = np.lib.test()
Running unit tests for numpy.lib
...
Ran 976 tests in 3.933s
OK
>>> result.errors
[]
>>> result.knownfail
[]
### Timer
class dipy.core.benchmarks.bench_sphere.Timer
Bases: object
Methods
duration_in_seconds
__init__(*args, **kwargs)
duration_in_seconds()
### bench_disperse_charges_alt
dipy.core.benchmarks.bench_sphere.bench_disperse_charges_alt()
### func_minimize_scipy
dipy.core.benchmarks.bench_sphere.func_minimize_scipy(init_pointset, num_iterations)
### cart2sphere
dipy.core.geometry.cart2sphere(x, y, z)
Return angles for Cartesian 3D coordinates x, y, and z
See doc for sphere2cart for angle conventions and derivation of the formulae.
$$0\le\theta\mathrm{(theta)}\le\pi$$ and $$-\pi\le\phi\mathrm{(phi)}\le\pi$$
Parameters:
xarray_like
x coordinate in Cartesian space
yarray_like
y coordinate in Cartesian space
zarray_like
z coordinate
Returns:
rarray
thetaarray
inclination (polar) angle
phiarray
azimuth angle
### cart_distance
dipy.core.geometry.cart_distance(pts1, pts2)
Cartesian distance between pts1 and pts2
If either of pts1 or pts2 is 2D, then we take the first dimension to index points, and the second indexes coordinate. More generally, we take the last dimension to be the coordinate dimension.
Parameters:
pts1(N,R) or (R,) array_like
where N is the number of points and R is the number of coordinates defining a point (R==3 for 3D)
pts2(N,R) or (R,) array_like
where N is the number of points and R is the number of coordinates defining a point (R==3 for 3D). It should be possible to broadcast pts1 against pts2
Returns:
d(N,) or (0,) array
Cartesian distances between corresponding points in pts1 and pts2
sphere_distance
distance between points on sphere surface
Examples
>>> cart_distance([0,0,0], [0,0,3])
3.0
a, b and c are 3-dimensional vectors which are the vertices of a triangle. The function returns the circumradius of the triangle, i.e the radius of the smallest circle that can contain the triangle. In the degenerate case when the 3 points are collinear it returns half the distance between the furthest apart points.
Parameters:
a, b, c(3,) array_like
the three vertices of the triangle
Returns:
### compose_matrix
dipy.core.geometry.compose_matrix(scale=None, shear=None, angles=None, translate=None, perspective=None)
Return 4x4 transformation matrix from sequence of transformations.
Code modified from the work of Christoph Gohlke link provided here http://www.lfd.uci.edu/~gohlke/code/transformations.py.html
This is the inverse of the decompose_matrix function.
Parameters:
scale(3,) array_like
Scaling factors.
sheararray_like
Shear factors for x-y, x-z, y-z axes.
anglesarray_like
Euler angles about static x, y, z axes.
translatearray_like
Translation vector along x, y, z axes.
perspectivearray_like
Perspective partition of matrix.
Returns:
matrix4x4 array
Examples
>>> import math
>>> import numpy as np
>>> import dipy.core.geometry as gm
>>> scale = np.random.random(3) - 0.5
>>> shear = np.random.random(3) - 0.5
>>> angles = (np.random.random(3) - 0.5) * (2*math.pi)
>>> trans = np.random.random(3) - 0.5
>>> persp = np.random.random(4) - 0.5
>>> M0 = gm.compose_matrix(scale, shear, angles, trans, persp)
### compose_transformations
dipy.core.geometry.compose_transformations(*mats)
Compose multiple 4x4 affine transformations in one 4x4 matrix
Parameters:
mat1array, (4, 4)
mat2array, (4, 4)
matNarray, (4, 4)
Returns:
matN x … x mat2 x mat1array, (4, 4)
### decompose_matrix
dipy.core.geometry.decompose_matrix(matrix)
Return sequence of transformations from transformation matrix.
Code modified from the excellent work of Christoph Gohlke link provided here: http://www.lfd.uci.edu/~gohlke/code/transformations.py.html
Parameters:
matrixarray_like
Non-degenerate homogeneous transformation matrix
Returns:
scale(3,) ndarray
Three scaling factors.
shear(3,) ndarray
Shear factors for x-y, x-z, y-z axes.
angles(3,) ndarray
Euler angles about static x, y, z axes.
translate(3,) ndarray
Translation vector along x, y, z axes.
perspectivendarray
Perspective partition of matrix.
Raises:
ValueError
If matrix is of wrong type or degenerate.
Examples
>>> import numpy as np
>>> T0=np.diag([2,1,1,1])
>>> scale, shear, angles, trans, persp = decompose_matrix(T0)
### dist_to_corner
dipy.core.geometry.dist_to_corner(affine)
Calculate the maximal distance from the center to a corner of a voxel, given an affine
Parameters:
affine4 by 4 array.
The spatial transformation from the measurement to the scanner space.
Returns:
dist: float
The maximal distance to the corner of a voxel, given voxel size encoded in the affine.
### euler_matrix
dipy.core.geometry.euler_matrix(ai, aj, ak, axes='sxyz')
Return homogeneous rotation matrix from Euler angles and axis sequence.
Code modified from the work of Christoph Gohlke link provided here http://www.lfd.uci.edu/~gohlke/code/transformations.py.html
Parameters:
ai, aj, akEuler’s roll, pitch and yaw angles
axesOne of 24 axis sequences as string or encoded tuple
Returns:
matrixndarray (4, 4)
Code modified from the work of Christoph Gohlke link provided here
http://www.lfd.uci.edu/~gohlke/code/transformations.py.html
Examples
>>> import numpy
>>> R = euler_matrix(1, 2, 3, 'syxz')
>>> numpy.allclose(numpy.sum(R[0]), -1.34786452)
True
>>> R = euler_matrix(1, 2, 3, (0, 1, 0, 1))
>>> numpy.allclose(numpy.sum(R[0]), -0.383436184)
True
>>> ai, aj, ak = (4.0*math.pi) * (numpy.random.random(3) - 0.5)
>>> for axes in _AXES2TUPLE.keys():
... _ = euler_matrix(ai, aj, ak, axes)
>>> for axes in _TUPLE2AXES.keys():
... _ = euler_matrix(ai, aj, ak, axes)
### is_hemispherical
dipy.core.geometry.is_hemispherical(vecs)
Test whether all points on a unit sphere lie in the same hemisphere.
Parameters:
vecsnumpy.ndarray
2D numpy array with shape (N, 3) where N is the number of points. All points must lie on the unit sphere.
Returns:
is_hemibool
If True, one can find a hemisphere that contains all the points. If False, then the points do not lie in any hemisphere
polenumpy.ndarray
If is_hemi == True, then pole is the “central” pole of the input vectors. Otherwise, pole is the zero vector.
References
### lambert_equal_area_projection_cart
dipy.core.geometry.lambert_equal_area_projection_cart(x, y, z)
Lambert Equal Area Projection from cartesian vector to plane
Return positions in $$(y_1,y_2)$$ plane corresponding to the directions of the vectors with cartesian coordinates xyz under the Lambert Equal Area Projection mapping (see Mardia and Jupp (2000), Directional Statistics, p. 161).
The Lambert EAP maps the upper hemisphere to the planar disc of radius 1 and the lower hemisphere to the planar annulus between radii 1 and 2, The Lambert EAP maps the upper hemisphere to the planar disc of radius 1 and the lower hemisphere to the planar annulus between radii 1 and 2. and vice versa.
See doc for sphere2cart for angle conventions
Parameters:
xarray_like
x coordinate in Cartesion space
yarray_like
y coordinate in Cartesian space
zarray_like
z coordinate
Returns:
y(N,2) array
planar coordinates of points following mapping by Lambert’s EAP.
### lambert_equal_area_projection_polar
dipy.core.geometry.lambert_equal_area_projection_polar(theta, phi)
Lambert Equal Area Projection from polar sphere to plane
Return positions in (y1,y2) plane corresponding to the points with polar coordinates (theta, phi) on the unit sphere, under the Lambert Equal Area Projection mapping (see Mardia and Jupp (2000), Directional Statistics, p. 161).
See doc for sphere2cart for angle conventions
• $$0 \le \theta \le \pi$$ and $$0 \le \phi \le 2 \pi$$
• $$|(y_1,y_2)| \le 2$$
The Lambert EAP maps the upper hemisphere to the planar disc of radius 1 and the lower hemisphere to the planar annulus between radii 1 and 2, and vice versa.
Parameters:
thetaarray_like
theta spherical coordinates
phiarray_like
phi spherical coordinates
Returns:
y(N,2) array
planar coordinates of points following mapping by Lambert’s EAP.
### nearest_pos_semi_def
dipy.core.geometry.nearest_pos_semi_def(B)
Least squares positive semi-definite tensor estimation
Parameters:
B(3,3) array_like
B matrix - symmetric. We do not check the symmetry.
Returns:
npds(3,3) array
Estimated nearest positive semi-definite array to matrix B.
References
[1]
Niethammer M, San Jose Estepar R, Bouix S, Shenton M, Westin CF. On diffusion tensor estimation. Conf Proc IEEE Eng Med Biol Soc. 2006;1:2622-5. PubMed PMID: 17946125; PubMed Central PMCID: PMC2791793.
Examples
>>> B = np.diag([1, 1, -1])
>>> nearest_pos_semi_def(B)
array([[ 0.75, 0. , 0. ],
[ 0. , 0.75, 0. ],
[ 0. , 0. , 0. ]])
### normalized_vector
dipy.core.geometry.normalized_vector(vec, axis=-1)
Return vector divided by its Euclidean (L2) norm
Parameters:
vecarray_like shape (3,)
Returns:
nvecarray shape (3,)
vector divided by L2 norm
Examples
>>> vec = [1, 2, 3]
>>> l2n = np.sqrt(np.dot(vec, vec))
>>> nvec = normalized_vector(vec)
>>> np.allclose(np.array(vec) / l2n, nvec)
True
>>> vec = np.array([[1, 2, 3]])
>>> vec.shape == (1, 3)
True
>>> normalized_vector(vec).shape == (1, 3)
True
### perpendicular_directions
dipy.core.geometry.perpendicular_directions(v, num=30, half=False)
Computes n evenly spaced perpendicular directions relative to a given vector v
Parameters:
varray (3,)
Array containing the three cartesian coordinates of vector v
numint, optional
Number of perpendicular directions to generate
halfbool, optional
If half is True, perpendicular directions are sampled on half of the unit circumference perpendicular to v, otherwive perpendicular directions are sampled on the full circumference. Default of half is False
Returns:
psamplesarray (n, 3)
array of vectors perpendicular to v
Notes
Perpendicular directions are estimated using the following two step procedure:
1) the perpendicular directions are first sampled in a unit circumference parallel to the plane normal to the x-axis.
2) Samples are then rotated and aligned to the plane normal to vector v. The rotational matrix for this rotation is constructed as reference frame basis which axis are the following:
• The first axis is vector v
• The second axis is defined as the normalized vector given by the
cross product between vector v and the unit vector aligned to the x-axis - The third axis is defined as the cross product between the previous computed vector and vector v.
Following this two steps, coordinates of the final perpendicular directions are given as:
$\left [ -\sin(a_{i}) \sqrt{{v_{y}}^{2}+{v_{z}}^{2}} \; , \; \frac{v_{x}v_{y}\sin(a_{i})-v_{z}\cos(a_{i})} {\sqrt{{v_{y}}^{2}+{v_{z}}^{2}}} \; , \; \frac{v_{x}v_{z}\sin(a_{i})-v_{y}\cos(a_{i})} {\sqrt{{v_{y}}^{2}+{v_{z}}^{2}}} \right ]$
This procedure has a singularity when vector v is aligned to the x-axis. To solve this singularity, perpendicular directions in procedure’s step 1 are defined in the plane normal to y-axis and the second axis of the rotated frame of reference is computed as the normalized vector given by the cross product between vector v and the unit vector aligned to the y-axis. Following this, the coordinates of the perpendicular directions are given as:
left [ -frac{left (v_{x}v_{y}sin(a_{i})+v_{z}cos(a_{i}) right )} {sqrt{{v_{x}}^{2}+{v_{z}}^{2}}} ; , ; sin(a_{i}) sqrt{{v_{x}}^{2}+{v_{z}}^{2}} ; , ; frac{v_{y}v_{z}sin(a_{i})+v_{x}cos(a_{i})} {sqrt{{v_{x}}^{2}+{v_{z}}^{2}}} right ]
For more details on this calculation, see here <http://gsoc2015dipydki.blogspot.it/2015/07/rnh-post-8-computing-perpendicular.html>_.
### rodrigues_axis_rotation
dipy.core.geometry.rodrigues_axis_rotation(r, theta)
Rodrigues formula
Rotation matrix for rotation around axis r for angle theta.
The rotation matrix is given by the Rodrigues formula:
R = Id + sin(theta)*Sn + (1-cos(theta))*Sn^2
with:
0 -nz ny
Sn = nz 0 -nx
-ny nx 0
where n = r / ||r||
In case the angle ||r|| is very small, the above formula may lead to numerical instabilities. We instead use a Taylor expansion around theta=0:
R = I + sin(theta)/tetha Sr + (1-cos(theta))/teta2 Sr^2
R = I + (1-theta2/6)*Sr + (1/2-theta2/24)*Sr^2
Parameters:
rarray_like shape (3,), axis
thetafloat, angle in degrees
Returns:
Rarray, shape (3,3), rotation matrix
Examples
>>> import numpy as np
>>> from dipy.core.geometry import rodrigues_axis_rotation
>>> v=np.array([0,0,1])
>>> u=np.array([1,0,0])
>>> R=rodrigues_axis_rotation(v,40)
>>> ur=np.dot(R,u)
40.0
### sph2latlon
dipy.core.geometry.sph2latlon(theta, phi)
Convert spherical coordinates to latitude and longitude.
Returns:
lat, lonndarray
Latitude and longitude.
### sphere2cart
dipy.core.geometry.sphere2cart(r, theta, phi)
Spherical to Cartesian coordinates
This is the standard physics convention where theta is the inclination (polar) angle, and phi is the azimuth angle.
Imagine a sphere with center (0,0,0). Orient it with the z axis running south-north, the y axis running west-east and the x axis from posterior to anterior. theta (the inclination angle) is the angle to rotate from the z-axis (the zenith) around the y-axis, towards the x axis. Thus the rotation is counter-clockwise from the point of view of positive y. phi (azimuth) gives the angle of rotation around the z-axis towards the y axis. The rotation is counter-clockwise from the point of view of positive z.
Equivalently, given a point P on the sphere, with coordinates x, y, z, theta is the angle between P and the z-axis, and phi is the angle between the projection of P onto the XY plane, and the X axis.
Geographical nomenclature designates theta as ‘co-latitude’, and phi as ‘longitude’
Parameters:
rarray_like
thetaarray_like
inclination or polar angle
phiarray_like
azimuth angle
Returns:
xarray
x coordinate(s) in Cartesion space
yarray
y coordinate(s) in Cartesian space
zarray
z coordinate
Notes
See these pages:
for excellent discussion of the many different conventions possible. Here we use the physics conventions, used in the wikipedia page.
Derivations of the formulae are simple. Consider a vector x, y, z of length r (norm of x, y, z). The inclination angle (theta) can be found from: cos(theta) == z / r -> z == r * cos(theta). This gives the hypotenuse of the projection onto the XY plane, which we will call Q. Q == r*sin(theta). Now x / Q == cos(phi) -> x == r * sin(theta) * cos(phi) and so on.
We have deliberately named this function sphere2cart rather than sph2cart to distinguish it from the Matlab function of that name, because the Matlab function uses an unusual convention for the angles that we did not want to replicate. The Matlab function is trivial to implement with the formulae given in the Matlab help.
### sphere_distance
Distance across sphere surface between pts1 and pts2
Parameters:
pts1(N,R) or (R,) array_like
where N is the number of points and R is the number of coordinates defining a point (R==3 for 3D)
pts2(N,R) or (R,) array_like
where N is the number of points and R is the number of coordinates defining a point (R==3 for 3D). It should be possible to broadcast pts1 against pts2
Radius of sphere. Default is to work out radius from mean of the length of each point vector
If True, check if the points are on the sphere surface - i.e check if the vector lengths in pts1 and pts2 are close to radius. Default is True.
Returns:
d(N,) or (0,) array
Distances between corresponding points in pts1 and pts2 across the spherical surface, i.e. the great circle distance
cart_distance
cartesian distance between points
vector_cosine
cosine of angle between vectors
Examples
>>> print('%.4f' % sphere_distance([0,1],[1,0]))
1.5708
>>> print('%.4f' % sphere_distance([0,3],[3,0]))
4.7124
### vec2vec_rotmat
dipy.core.geometry.vec2vec_rotmat(u, v)
rotation matrix from 2 unit vectors
u, v being unit 3d vectors return a 3x3 rotation matrix R than aligns u to v.
In general there are many rotations that will map u to v. If S is any rotation using v as an axis then R.S will also map u to v since (S.R)u = S(Ru) = Sv = v. The rotation R returned by vec2vec_rotmat leaves fixed the perpendicular to the plane spanned by u and v.
The transpose of R will align v to u.
Parameters:
uarray, shape(3,)
varray, shape(3,)
Returns:
Rarray, shape(3,3)
Examples
>>> import numpy as np
>>> from dipy.core.geometry import vec2vec_rotmat
>>> u=np.array([1,0,0])
>>> v=np.array([0,1,0])
>>> R=vec2vec_rotmat(u,v)
>>> np.dot(R,u)
array([ 0., 1., 0.])
>>> np.dot(R.T,v)
array([ 1., 0., 0.])
### vector_cosine
dipy.core.geometry.vector_cosine(vecs1, vecs2)
Cosine of angle between two (sets of) vectors
The cosine of the angle between two vectors v1 and v2 is given by the inner product of v1 and v2 divided by the product of the vector lengths:
v_cos = np.inner(v1, v2) / (np.sqrt(np.sum(v1**2)) *
np.sqrt(np.sum(v2**2)))
Parameters:
vecs1(N, R) or (R,) array_like
N vectors (as rows) or single vector. Vectors have R elements.
vecs1(N, R) or (R,) array_like
N vectors (as rows) or single vector. Vectors have R elements. It should be possible to broadcast vecs1 against vecs2
Returns:
vcos(N,) or (0,) array
Vector cosines. To get the angles you will need np.arccos
Notes
The vector cosine will be the same as the correlation only if all the input vectors have zero mean.
### vector_norm
dipy.core.geometry.vector_norm(vec, axis=-1, keepdims=False)
Return vector Euclidean (L2) norm
Parameters:
vecarray_like
Vectors to norm.
axisint
Axis over which to norm. By default norm over last axis. If axis is None, vec is flattened then normed.
keepdimsbool
If True, the output will have the same number of dimensions as vec, with shape 1 on axis.
Returns:
normarray
Euclidean norms of vectors.
Examples
>>> import numpy as np
>>> vec = [[8, 15, 0], [0, 36, 77]]
>>> vector_norm(vec)
array([ 17., 85.])
>>> vector_norm(vec, keepdims=True)
array([[ 17.],
[ 85.]])
>>> vector_norm(vec, axis=0)
array([ 8., 39., 77.])
### GradientTable
Bases: object
Parameters:
Diffusion gradients. The direction of each of these vectors corresponds to the b-vector, and the length corresponds to the b-value.
b0_thresholdfloat
Gradients with b-value less than or equal to b0_threshold are considered as b0s i.e. without diffusion weighting.
Notes
The GradientTable object is immutable. Do NOT assign attributes. If you have your gradient table in a bval & bvec format, we recommend using the factory function gradient_table
Attributes:
bvals(N,) ndarray
The b-value, or magnitude, of each gradient direction.
qvals: (N,) ndarray
The q-value for each gradient direction. Needs big and small delta.
bvecs(N,3) ndarray
The direction, represented as a unit vector, of each gradient.
Boolean array indicating which gradients have no diffusion weighting, ie b-value is close to 0.
b0_thresholdfloat
Gradients with b-value less than or equal to b0_threshold are considered to not have diffusion weighting.
btens(N,3,3) ndarray
The b-tensor of each gradient direction.
Methods
bvals()
bvecs()
property info
qvals()
tau()
### HemiSphere
class dipy.core.gradients.HemiSphere(x=None, y=None, z=None, theta=None, phi=None, xyz=None, faces=None, edges=None, tol=1e-05)
Bases: Sphere
Points on the unit sphere.
A HemiSphere is similar to a Sphere but it takes antipodal symmetry into account. Antipodal symmetry means that point v on a HemiSphere is the same as the point -v. Duplicate points are discarded when constructing a HemiSphere (including antipodal duplicates). edges and faces are remapped to the remaining points as closely as possible.
The HemiSphere can be constructed using one of three conventions:
HemiSphere(x, y, z)
HemiSphere(xyz=xyz)
HemiSphere(theta=theta, phi=phi)
Parameters:
x, y, z1-D array_like
Vertices as x-y-z coordinates.
theta, phi1-D array_like
Vertices as spherical coordinates. Theta and phi are the inclination and azimuth angles respectively.
xyz(N, 3) ndarray
Vertices as x-y-z coordinates.
faces(N, 3) ndarray
Indices into vertices that form triangular faces. If unspecified, the faces are computed using a Delaunay triangulation.
edges(N, 2) ndarray
Edges between vertices. If unspecified, the edges are derived from the faces.
tolfloat
Angle in degrees. Vertices that are less than tol degrees apart are treated as duplicates.
Sphere
Attributes:
x
y
z
Methods
Find the index of the vertex in the Sphere closest to the input vector, taking into account antipodal symmetry from_sphere(sphere[, tol]) Create instance from a Sphere Create a full Sphere from a HemiSphere subdivide([n]) Create a more subdivided HemiSphere
edges faces vertices
__init__(x=None, y=None, z=None, theta=None, phi=None, xyz=None, faces=None, edges=None, tol=1e-05)
Create a HemiSphere from points
faces()
find_closest(xyz)
Find the index of the vertex in the Sphere closest to the input vector, taking into account antipodal symmetry
Parameters:
xyzarray-like, 3 elements
A unit vector
Returns:
idxint
The index into the Sphere.vertices array that gives the closest vertex (in angle).
classmethod from_sphere(sphere, tol=1e-05)
Create instance from a Sphere
mirror()
Create a full Sphere from a HemiSphere
subdivide(n=1)
Create a more subdivided HemiSphere
See Sphere.subdivide for full documentation.
### auto_attr
Decorator to create OneTimeProperty attributes.
Parameters:
funcmethod
The method that will be called the first time to compute a value. Afterwards, the method’s name will be a standard attribute holding the value of this computation.
Examples
>>> class MagicProp(object):
... @auto_attr
... def a(self):
... return 99
...
>>> x = MagicProp()
>>> 'a' in x.__dict__
False
>>> x.a
99
>>> 'a' in x.__dict__
True
### btens_to_params
Compute trace, anisotropy and assymetry parameters from b-tensors.
Parameters:
btens(3, 3) OR (N, 3, 3) numpy.ndarray
input b-tensor, or b-tensors, where N = number of b-tensors
ztolfloat
Any parameters smaller than this value are considered to be 0
Returns:
bval: numpy.ndarray
b-value(s) (trace(s))
bdelta: numpy.ndarray
normalized tensor anisotropy(s)
b_eta: numpy.ndarray
tensor assymetry(s)
Notes
This function can be used to get b-tensor parameters directly from the GradientTable btens attribute.
Examples
>>> lte = np.array([[1, 0, 0], [0, 0, 0], [0, 0, 0]])
>>> bval, bdelta, b_eta = btens_to_params(lte)
>>> print("bval={}; bdelta={}; b_eta={}".format(bdelta, bval, b_eta))
bval=[ 1.]; bdelta=[ 1.]; b_eta=[ 0.]
### check_multi_b
Parameters:
n_bvalsint
The number of different b-values you are checking for.
non_zerobool
Whether to check only non-zero bvalues. In this case, we will require at least n_bvals non-zero b-values (where non-zero is defined depending on the gtab object’s b0_threshold attribute)
bmagint
The order of magnitude of the b-values used. The function will normalize the b-values relative $$10^{bmag}$$. Default: derive this value from the maximal b-value provided: $$bmag=log_{10}(max(bvals)) - 1$$.
Returns:
boolWhether there are at least n_bvals different b-values in the
### deprecate_with_version
dipy.core.gradients.deprecate_with_version(message, since='', until='', version_comparator=<function cmp_pkg_version>, warn_class=<class 'DeprecationWarning'>, error_class=<class 'dipy.utils.deprecator.ExpiredDeprecationError'>)
Return decorator function function for deprecation warning / error.
The decorated function / method will:
• Raise the given warning_class warning when the function / method gets called, up to (and including) version until (if specified);
• Raise the given error_class error when the function / method gets called, when the package version is greater than version until (if specified).
Parameters:
messagestr
Message explaining deprecation, giving possible alternatives.
sincestr, optional
Released version at which object was first deprecated.
untilstr, optional
Last released version at which this function will still raise a deprecation warning. Versions higher than this will raise an error.
version_comparatorcallable
Callable accepting string as argument, and return 1 if string represents a higher version than encoded in the version_comparator, 0 if the version is equal, and -1 if the version is lower. For example, the version_comparator may compare the input version string to the current package version string.
warn_classclass, optional
Class of warning to generate for deprecation.
error_classclass, optional
Class of error to generate when version_comparator returns 1 for a given argument of until.
Returns:
deprecatorfunc
Function returning a decorator.
### disperse_charges
Models electrostatic repulsion on the unit sphere
Places charges on a sphere and simulates the repulsive forces felt by each one. Allows the charges to move for some number of iterations and returns their final location as well as the total potential of the system at each step.
Parameters:
hemiHemiSphere
Points on a unit sphere.
itersint
Number of iterations to run.
constfloat
Using a smaller const could provide a more accurate result, but will need more iterations to converge.
Returns:
hemiHemiSphere
Distributed points on a unit sphere.
potentialndarray
The electrostatic potential at each iteration. This can be useful to check if the repulsion converged to a minimum.
Notes
This function is meant to be used with diffusion imaging so antipodal symmetry is assumed. Therefore, each charge must not only be unique, but if there is a charge at +x, there cannot be a charge at -x. These are treated as the same location and because the distance between the two charges will be zero, the result will be unstable.
### generate_bvecs
Generates N bvectors.
Uses dipy.core.sphere.disperse_charges to model electrostatic repulsion on a unit sphere.
Parameters:
Nint
The number of bvectors to generate. This should be equal to the number of bvals used.
itersint
Number of iterations to run.
Returns:
bvecs(N,3) ndarray
The generated directions, represented as a unit vector, of each gradient.
### get_bval_indices
Get indices where the b-value is bval
Parameters:
bvals: ndarray
Array containing the b-values
bval: float or int
b-value to extract indices
tol: int
The tolerated gap between the b-values to extract and the actual b-values.
Returns:
Array of indices where the b-value is bval
A general function for creating diffusion MR gradients.
It reads, loads and prepares scanner parameters like the b-values and b-vectors so that they can be useful during the reconstruction process.
Parameters:
bvalscan be any of the four options
1. an array of shape (N,) or (1, N) or (N, 1) with the b-values.
2. a path for the file which contains an array like the above (1).
3. an array of shape (N, 4) or (4, N). Then this parameter is considered to be a b-table which contains both bvals and bvecs. In this case the next parameter is skipped.
4. a path for the file which contains an array like the one at (3).
bvecscan be any of two options
1. an array of shape (N, 3) or (3, N) with the b-vectors.
2. a path for the file which contains an array like the previous.
big_deltafloat
acquisition pulse separation time in seconds (default None)
small_deltafloat
acquisition pulse duration time in seconds (default None)
b0_thresholdfloat
All b-values with values less than or equal to bo_threshold are considered as b0s i.e. without diffusion weighting.
atolfloat
All b-vectors need to be unit vectors up to a tolerance.
btenscan be any of three options
1. a string specifying the shape of the encoding tensor for all volumes in data. Options: ‘LTE’, ‘PTE’, ‘STE’, ‘CTE’ corresponding to linear, planar, spherical, and “cigar-shaped” tensor encoding. Tensors are rotated so that linear and cigar tensors are aligned with the corresponding gradient direction and the planar tensor’s normal is aligned with the corresponding gradient direction. Magnitude is scaled to match the b-value.
2. an array of strings of shape (N,), (N, 1), or (1, N) specifying encoding tensor shape for each volume separately. N corresponds to the number volumes in data. Options for elements in array: ‘LTE’, ‘PTE’, ‘STE’, ‘CTE’ corresponding to linear, planar, spherical, and “cigar-shaped” tensor encoding. Tensors are rotated so that linear and cigar tensors are aligned with the corresponding gradient direction and the planar tensor’s normal is aligned with the corresponding gradient direction. Magnitude is scaled to match the b-value.
3. an array of shape (N,3,3) specifying the b-tensor of each volume exactly. N corresponds to the number volumes in data. No rotation or scaling is performed.
Returns:
Notes
1. Often b0s (b-values which correspond to images without diffusion weighting) have 0 values however in some cases the scanner cannot provide b0s of an exact 0 value and it gives a bit higher values e.g. 6 or 12. This is the purpose of the b0_threshold in the __init__.
2. We assume that the minimum number of b-values is 7.
3. B-vectors should be unit vectors.
Examples
>>> from dipy.core.gradients import gradient_table
>>> bvals = 1500 * np.ones(7)
>>> bvals[0] = 0
>>> sq2 = np.sqrt(2) / 2
>>> bvecs = np.array([[0, 0, 0],
... [1, 0, 0],
... [0, 1, 0],
... [0, 0, 1],
... [sq2, sq2, 0],
... [sq2, 0, sq2],
... [0, sq2, sq2]])
>>> gt.bvecs.shape == bvecs.shape
True
>>> gt.bvecs.shape == bvecs.T.shape
False
Creates a GradientTable from a bvals array and a bvecs array
Parameters:
bvalsarray_like (N,)
The b-value, or magnitude, of each gradient direction.
bvecsarray_like (N, 3)
The direction, represented as a unit vector, of each gradient.
b0_thresholdfloat
Gradients with b-value less than or equal to bo_threshold are considered to not have diffusion weighting.
atolfloat
Each vector in bvecs must be a unit vectors up to a tolerance of atol.
btenscan be any of three options
1. a string specifying the shape of the encoding tensor for all volumes in data. Options: ‘LTE’, ‘PTE’, ‘STE’, ‘CTE’ corresponding to linear, planar, spherical, and “cigar-shaped” tensor encoding. Tensors are rotated so that linear and cigar tensors are aligned with the corresponding gradient direction and the planar tensor’s normal is aligned with the corresponding gradient direction. Magnitude is scaled to match the b-value.
2. an array of strings of shape (N,), (N, 1), or (1, N) specifying encoding tensor shape for each volume separately. N corresponds to the number volumes in data. Options for elements in array: ‘LTE’, ‘PTE’, ‘STE’, ‘CTE’ corresponding to linear, planar, spherical, and “cigar-shaped” tensor encoding. Tensors are rotated so that linear and cigar tensors are aligned with the corresponding gradient direction and the planar tensor’s normal is aligned with the corresponding gradient direction. Magnitude is scaled to match the b-value.
3. an array of shape (N,3,3) specifying the b-tensor of each volume exactly. N corresponds to the number volumes in data. No rotation or scaling is performed.
Returns:
Other Parameters:
**kwargsdict
Other keyword inputs are passed to GradientTable.
A general function for creating diffusion MR gradients.
It reads, loads and prepares scanner parameters like the b-values and b-vectors so that they can be useful during the reconstruction process.
Parameters:
bvecscan be any of two options
1. an array of shape (N, 3) or (3, N) with the b-vectors.
2. a path for the file which contains an array like the previous.
big_deltafloat or array of shape (N,)
acquisition pulse separation time in seconds
small_deltafloat
acquisition pulse duration time in seconds
b0_thresholdfloat
All b-values with values less than or equal to bo_threshold are considered as b0s i.e. without diffusion weighting.
atolfloat
All b-vectors need to be unit vectors up to a tolerance.
Returns:
Notes
1. Often b0s (b-values which correspond to images without diffusion weighting) have 0 values however in some cases the scanner cannot provide b0s of an exact 0 value and it gives a bit higher values e.g. 6 or 12. This is the purpose of the b0_threshold in the __init__.
2. We assume that the minimum number of b-values is 7.
3. B-vectors should be unit vectors.
Examples
>>> from dipy.core.gradients import (
>>> gradient_strength = .03e-3 * np.ones(7) # clinical strength at 30 mT/m
>>> big_delta = .03 # pulse separation of 30ms
>>> small_delta = 0.01 # pulse duration of 10ms
>>> sq2 = np.sqrt(2) / 2
>>> bvecs = np.array([[0, 0, 0],
... [1, 0, 0],
... [0, 1, 0],
... [0, 0, 1],
... [sq2, sq2, 0],
... [sq2, 0, sq2],
... [0, sq2, sq2]])
A general function for creating diffusion MR gradients.
It reads, loads and prepares scanner parameters like the b-values and b-vectors so that they can be useful during the reconstruction process.
Parameters:
qvalsan array of shape (N,),
q-value given in 1/mm
bvecscan be any of two options
1. an array of shape (N, 3) or (3, N) with the b-vectors.
2. a path for the file which contains an array like the previous.
big_deltafloat or array of shape (N,)
acquisition pulse separation time in seconds
small_deltafloat
acquisition pulse duration time in seconds
b0_thresholdfloat
All b-values with values less than or equal to bo_threshold are considered as b0s i.e. without diffusion weighting.
atolfloat
All b-vectors need to be unit vectors up to a tolerance.
Returns:
Notes
1. Often b0s (b-values which correspond to images without diffusion weighting) have 0 values however in some cases the scanner cannot provide b0s of an exact 0 value and it gives a bit higher values e.g. 6 or 12. This is the purpose of the b0_threshold in the __init__.
2. We assume that the minimum number of b-values is 7.
3. B-vectors should be unit vectors.
Examples
>>> from dipy.core.gradients import gradient_table_from_qvals_bvecs
>>> qvals = 30. * np.ones(7)
>>> big_delta = .03 # pulse separation of 30ms
>>> small_delta = 0.01 # pulse duration of 10ms
>>> qvals[0] = 0
>>> sq2 = np.sqrt(2) / 2
>>> bvecs = np.array([[0, 0, 0],
... [1, 0, 0],
... [0, 1, 0],
... [0, 0, 1],
... [sq2, sq2, 0],
... [sq2, 0, sq2],
... [0, sq2, sq2]])
... big_delta, small_delta)
### inv
Compute the inverse of a matrix.
Parameters:
aarray_like
Square matrix to be inverted.
overwrite_abool, optional
Discard data in a (may improve performance). Default is False.
check_finitebool, optional
Whether to check that the input matrix contains only finite numbers. Disabling may give a performance gain, but may result in problems (crashes, non-termination) if the inputs do contain infinities or NaNs.
Returns:
ainvndarray
Inverse of the matrix a.
Raises:
LinAlgError
If a is singular.
ValueError
If a is not square, or not 2D.
Examples
>>> import numpy as np
>>> from scipy import linalg
>>> a = np.array([[1., 2.], [3., 4.]])
>>> linalg.inv(a)
array([[-2. , 1. ],
[ 1.5, -0.5]])
>>> np.dot(a, linalg.inv(a))
array([[ 1., 0.],
[ 0., 1.]])
### orientation_from_string
Return an array representation of an ornt string.
### orientation_to_string
Return a string representation of a 3d ornt.
### ornt_mapping
Calculate the mapping needing to get from orn1 to orn2.
### params_to_btens
Compute b-tensor from trace, anisotropy and assymetry parameters.
Parameters:
bval: int or float
b-value (>= 0)
bdelta: int or float
normalized tensor anisotropy (>= -0.5 and <= 1)
b_eta: int or float
tensor assymetry (>= 0 and <= 1)
Returns:
(3, 3) numpy.ndarray
output b-tensor
Notes
Implements eq. 7.11. p. 231 in [1].
References
[1]
1. Topgaard, NMR methods for studying microscopic diffusion
anisotropy, in: R. Valiullin (Ed.), Diffusion NMR of Confined Systems: Fluid Transport in Porous Solids and Heterogeneous Materials, Royal Society of Chemistry, Cambridge, UK, 2016.
### polar
Compute the polar decomposition.
Returns the factors of the polar decomposition [1] u and p such that a = up (if side is “right”) or a = pu (if side is “left”), where p is positive semidefinite. Depending on the shape of a, either the rows or columns of u are orthonormal. When a is a square array, u is a square unitary array. When a is not square, the “canonical polar decomposition” [2] is computed.
Parameters:
a(m, n) array_like
The array to be factored.
side{‘left’, ‘right’}, optional
Determines whether a right or left polar decomposition is computed. If side is “right”, then a = up. If side is “left”, then a = pu. The default is “right”.
Returns:
u(m, n) ndarray
If a is square, then u is unitary. If m > n, then the columns of a are orthonormal, and if m < n, then the rows of u are orthonormal.
pndarray
p is Hermitian positive semidefinite. If a is nonsingular, p is positive definite. The shape of p is (n, n) or (m, m), depending on whether side is “right” or “left”, respectively.
References
[1]
R. A. Horn and C. R. Johnson, “Matrix Analysis”, Cambridge University Press, 1985.
[2]
N. J. Higham, “Functions of Matrices: Theory and Computation”, SIAM, 2008.
Examples
>>> import numpy as np
>>> from scipy.linalg import polar
>>> a = np.array([[1, -1], [2, 4]])
>>> u, p = polar(a)
>>> u
array([[ 0.85749293, -0.51449576],
[ 0.51449576, 0.85749293]])
>>> p
array([[ 1.88648444, 1.2004901 ],
[ 1.2004901 , 3.94446746]])
A non-square example, with m < n:
>>> b = np.array([[0.5, 1, 2], [1.5, 3, 4]])
>>> u, p = polar(b)
>>> u
array([[-0.21196618, -0.42393237, 0.88054056],
[ 0.39378971, 0.78757942, 0.4739708 ]])
>>> p
array([[ 0.48470147, 0.96940295, 1.15122648],
[ 0.96940295, 1.9388059 , 2.30245295],
[ 1.15122648, 2.30245295, 3.65696431]])
>>> u.dot(p) # Verify the decomposition.
array([[ 0.5, 1. , 2. ],
[ 1.5, 3. , 4. ]])
>>> u.dot(u.T) # The rows of u are orthonormal.
array([[ 1.00000000e+00, -2.07353665e-17],
[ -2.07353665e-17, 1.00000000e+00]])
Another non-square example, with m > n:
>>> c = b.T
>>> u, p = polar(c)
>>> u
array([[-0.21196618, 0.39378971],
[-0.42393237, 0.78757942],
[ 0.88054056, 0.4739708 ]])
>>> p
array([[ 1.23116567, 1.93241587],
[ 1.93241587, 4.84930602]])
>>> u.dot(p) # Verify the decomposition.
array([[ 0.5, 1.5],
[ 1. , 3. ],
[ 2. , 4. ]])
>>> u.T.dot(u) # The columns of u are orthonormal.
array([[ 1.00000000e+00, -1.26363763e-16],
[ -1.26363763e-16, 1.00000000e+00]])
### reorient_bvecs
Reorient the directions in a GradientTable.
When correcting for motion, rotation of the diffusion-weighted volumes might cause systematic bias in rotationally invariant measures, such as FA and MD, and also cause characteristic biases in tractography, unless the gradient directions are appropriately reoriented to compensate for this effect [Leemans2009].
Parameters:
The nominal gradient table with which the data were acquired.
affineslist or ndarray of shape (n, 4, 4) or (n, 3, 3)
Each entry in this list or array contain either an affine transformation (4,4) or a rotation matrix (3, 3). In both cases, the transformations encode the rotation that was applied to the image corresponding to one of the non-zero gradient directions (ordered according to their order in gtab.bvecs[~gtab.b0s_mask])
Returns:
gtaba GradientTable class instance with the reoriented directions
References
The B-Matrix Must Be Rotated When Correcting for Subject Motion in DTI Data. Leemans, A. and Jones, D.K. (2009). MRM, 61: 1336-1349
### reorient_vectors
Change the orientation of gradients or other vectors.
Moves vectors, storted along axis, from current_ornt to new_ornt. For example the vector [x, y, z] in “RAS” will be [-x, -y, z] in “LPS”.
R: Right A: Anterior S: Superior L: Left P: Posterior I: Inferior
### round_bvals
“This function rounds the b-values
Parameters:
bvalsndarray
Array containing the b-values
bmagint
The order of magnitude to round the b-values. If not given b-values will be rounded relative to the order of magnitude $$bmag = (bmagmax - 1)$$, where bmaxmag is the magnitude order of the larger b-value.
Returns:
rbvalsndarray
Array containing the rounded b-values
### unique_bvals
This function gives the unique rounded b-values of the data
• deprecated from version: 1.2
• Raises <class ‘dipy.utils.deprecator.ExpiredDeprecationError’> as of version: 1.4
Parameters:
bvalsndarray
Array containing the b-values
bmagint
The order of magnitude that the bvalues have to differ to be considered an unique b-value. B-values are also rounded up to this order of magnitude. Default: derive this value from the maximal b-value provided: $$bmag=log_{10}(max(bvals)) - 1$$.
rbvalsbool, optional
If True function also returns all individual rounded b-values. Default: False
Returns:
ubvalsndarray
Array containing the rounded unique b-values
### unique_bvals_magnitude
This function gives the unique rounded b-values of the data
Parameters:
bvalsndarray
Array containing the b-values
bmagint
The order of magnitude that the bvalues have to differ to be considered an unique b-value. B-values are also rounded up to this order of magnitude. Default: derive this value from the maximal b-value provided: $$bmag=log_{10}(max(bvals)) - 1$$.
rbvalsbool, optional
If True function also returns all individual rounded b-values. Default: False
Returns:
ubvalsndarray
Array containing the rounded unique b-values
### unique_bvals_tolerance
Gives the unique b-values of the data, within a tolerance gap
The b-values must be regrouped in clusters easily separated by a distance greater than the tolerance gap. If all the b-values of a cluster fit within the tolerance gap, the highest b-value is kept.
Parameters:
bvalsndarray
Array containing the b-values
tolint
The tolerated gap between the b-values to extract and the actual b-values.
Returns:
ubvalsndarray
Array containing the unique b-values using the median value for each cluster
### vec2vec_rotmat
rotation matrix from 2 unit vectors
u, v being unit 3d vectors return a 3x3 rotation matrix R than aligns u to v.
In general there are many rotations that will map u to v. If S is any rotation using v as an axis then R.S will also map u to v since (S.R)u = S(Ru) = Sv = v. The rotation R returned by vec2vec_rotmat leaves fixed the perpendicular to the plane spanned by u and v.
The transpose of R will align v to u.
Parameters:
uarray, shape(3,)
varray, shape(3,)
Returns:
Rarray, shape(3,3)
Examples
>>> import numpy as np
>>> from dipy.core.geometry import vec2vec_rotmat
>>> u=np.array([1,0,0])
>>> v=np.array([0,1,0])
>>> R=vec2vec_rotmat(u,v)
>>> np.dot(R,u)
array([ 0., 1., 0.])
>>> np.dot(R.T,v)
array([ 1., 0., 0.])
### vector_norm
Return vector Euclidean (L2) norm
Parameters:
vecarray_like
Vectors to norm.
axisint
Axis over which to norm. By default norm over last axis. If axis is None, vec is flattened then normed.
keepdimsbool
If True, the output will have the same number of dimensions as vec, with shape 1 on axis.
Returns:
normarray
Euclidean norms of vectors.
Examples
>>> import numpy as np
>>> vec = [[8, 15, 0], [0, 36, 77]]
>>> vector_norm(vec)
array([ 17., 85.])
>>> vector_norm(vec, keepdims=True)
array([[ 17.],
[ 85.]])
>>> vector_norm(vec, axis=0)
array([ 8., 39., 77.])
### warn
Issue a warning, or maybe ignore it or raise an exception.
### Graph
class dipy.core.graph.Graph
Bases: object
A simple graph class
Methods
add_edge add_node all_paths children del_node del_node_and_edges down down_short parents shortest_path up up_short
__init__()
A graph class with nodes and edges :-)
This class allows us to:
1. find the shortest path
2. find all paths
4. get parent & children nodes
Examples
>>> from dipy.core.graph import Graph
>>> g=Graph()
>>> g.up_short('d')
['d', 'b', 'a']
all_paths(graph, start, end=None, path=None)
children(n)
del_node(n)
del_node_and_edges(n)
down(n)
down_short(n)
parents(n)
shortest_path(graph, start, end=None, path=None)
up(n)
up_short(n)
### histeq
dipy.core.histeq.histeq(arr, num_bins=256)
Performs an histogram equalization on arr. This was taken from: http://www.janeriksolem.net/2009/06/histogram-equalization-with-python-and.html
Parameters:
arrndarray
Image on which to perform histogram equalization.
num_binsint
Number of bins used to construct the histogram.
Returns:
resultndarray
Histogram equalized image.
### Interpolator
class dipy.core.interpolation.Interpolator(data, voxel_size)
Bases: object
Class to be subclassed by different interpolator types
__init__(data, voxel_size)
### NearestNeighborInterpolator
class dipy.core.interpolation.NearestNeighborInterpolator(data, voxel_size)
Bases: Interpolator
Interpolates data using nearest neighbor interpolation
__init__(data, voxel_size)
### OutsideImage
class dipy.core.interpolation.OutsideImage
Bases: Exception
Attributes:
args
Methods
with_traceback Exception.with_traceback(tb) -- set self.__traceback__ to tb and return self.
__init__(*args, **kwargs)
### TriLinearInterpolator
class dipy.core.interpolation.TriLinearInterpolator(data, voxel_size)
Bases: Interpolator
Interpolates data using trilinear interpolation
interpolate 4d diffusion volume using 3 indices, ie data[x, y, z]
__init__(data, voxel_size)
### interp_rbf
dipy.core.interpolation.interp_rbf()
Interpolate data on the sphere, using radial basis functions.
Parameters:
data(N,) ndarray
Function values on the unit sphere.
sphere_originSphere
Positions of data values.
sphere_targetSphere
M target positions for which to interpolate.
epsilonfloat
Radial basis function spread parameter. Defaults to approximate average distance between nodes.
a good start
smoothfloat
values greater than zero increase the smoothness of the approximation with 0 as pure interpolation. Default: 0.1
normstr
A string indicating the function that returns the “distance” between two points. ‘angle’ - The angle between two vectors ‘euclidean_norm’ - The Euclidean distance
Returns:
v(M,) ndarray
Interpolated values.
scipy.interpolate.Rbf
### interpolate_scalar_2d
dipy.core.interpolation.interpolate_scalar_2d(image, locations)
Bilinear interpolation of a 2D scalar image
Interpolates the 2D image at the given locations. This function is a wrapper for _interpolate_scalar_2d for testing purposes, it is equivalent to scipy.ndimage.map_coordinates with bilinear interpolation
Parameters:
fieldarray, shape (S, R)
the 2D image to be interpolated
locationsarray, shape (n, 2)
(locations[i,0], locations[i,1]), 0<=i<n must contain the row and column coordinates to interpolate the image at
Returns:
outarray, shape (n,)
out[i], 0<=i<n will be the interpolated scalar at coordinates locations[i,:], or 0 if locations[i,:] is outside the image
insidearray, (n,)
if locations[i:] is inside the image then inside[i]=1, else inside[i]=0
### interpolate_scalar_3d
dipy.core.interpolation.interpolate_scalar_3d(image, locations)
Trilinear interpolation of a 3D scalar image
Interpolates the 3D image at the given locations. This function is a wrapper for _interpolate_scalar_3d for testing purposes, it is equivalent to scipy.ndimage.map_coordinates with trilinear interpolation
Parameters:
fieldarray, shape (S, R, C)
the 3D image to be interpolated
locationsarray, shape (n, 3)
(locations[i,0], locations[i,1], locations[i,2), 0<=i<n must contain the coordinates to interpolate the image at
Returns:
outarray, shape (n,)
out[i], 0<=i<n will be the interpolated scalar at coordinates locations[i,:], or 0 if locations[i,:] is outside the image
insidearray, (n,)
if locations[i,:] is inside the image then inside[i]=1, else inside[i]=0
### interpolate_scalar_nn_2d
dipy.core.interpolation.interpolate_scalar_nn_2d(image, locations)
Nearest neighbor interpolation of a 2D scalar image
Interpolates the 2D image at the given locations. This function is a wrapper for _interpolate_scalar_nn_2d for testing purposes, it is equivalent to scipy.ndimage.map_coordinates with nearest neighbor interpolation
Parameters:
imagearray, shape (S, R)
the 2D image to be interpolated
locationsarray, shape (n, 2)
(locations[i,0], locations[i,1]), 0<=i<n must contain the row and column coordinates to interpolate the image at
Returns:
outarray, shape (n,)
out[i], 0<=i<n will be the interpolated scalar at coordinates locations[i,:], or 0 if locations[i,:] is outside the image
insidearray, (n,)
if locations[i:] is inside the image then inside[i]=1, else inside[i]=0
### interpolate_scalar_nn_3d
dipy.core.interpolation.interpolate_scalar_nn_3d(image, locations)
Nearest neighbor interpolation of a 3D scalar image
Interpolates the 3D image at the given locations. This function is a wrapper for _interpolate_scalar_nn_3d for testing purposes, it is equivalent to scipy.ndimage.map_coordinates with nearest neighbor interpolation
Parameters:
imagearray, shape (S, R, C)
the 3D image to be interpolated
locationsarray, shape (n, 3)
(locations[i,0], locations[i,1], locations[i,2), 0<=i<n must contain the coordinates to interpolate the image at
Returns:
outarray, shape (n,)
out[i], 0<=i<n will be the interpolated scalar at coordinates locations[i,:], or 0 if locations[i,:] is outside the image
insidearray, (n,)
if locations[i,:] is inside the image then inside[i]=1, else inside[i]=0
### interpolate_vector_2d
dipy.core.interpolation.interpolate_vector_2d(field, locations)
Bilinear interpolation of a 2D vector field
Interpolates the 2D vector field at the given locations. This function is a wrapper for _interpolate_vector_2d for testing purposes, it is equivalent to using scipy.ndimage.map_coordinates with bilinear interpolation at each vector component
Parameters:
fieldarray, shape (S, R, 2)
the 2D vector field to be interpolated
locationsarray, shape (n, 2)
(locations[i,0], locations[i,1]), 0<=i<n must contain the row and column coordinates to interpolate the vector field at
Returns:
outarray, shape (n, 2)
out[i,:], 0<=i<n will be the interpolated vector at coordinates locations[i,:], or (0,0) if locations[i,:] is outside the field
insidearray, (n,)
if (locations[i,0], locations[i,1]) is inside the vector field then inside[i]=1, else inside[i]=0
### interpolate_vector_3d
dipy.core.interpolation.interpolate_vector_3d(field, locations)
Trilinear interpolation of a 3D vector field
Interpolates the 3D vector field at the given locations. This function is a wrapper for _interpolate_vector_3d for testing purposes, it is equivalent to using scipy.ndimage.map_coordinates with trilinear interpolation at each vector component
Parameters:
fieldarray, shape (S, R, C, 3)
the 3D vector field to be interpolated
locationsarray, shape (n, 3)
(locations[i,0], locations[i,1], locations[i,2), 0<=i<n must contain the coordinates to interpolate the vector field at
Returns:
outarray, shape (n, 3)
out[i,:], 0<=i<n will be the interpolated vector at coordinates locations[i,:], or (0,0,0) if locations[i,:] is outside the field
insidearray, (n,)
if locations[i,:] is inside the vector field then inside[i]=1, else inside[i]=0
### map_coordinates_trilinear_iso
dipy.core.interpolation.map_coordinates_trilinear_iso()
Trilinear interpolation (isotropic voxel size)
Has similar behavior to map_coordinates from scipy.ndimage
Parameters:
dataarray, float64 shape (X, Y, Z)
pointsarray, float64 shape(N, 3)
data_stridesarray npy_intp shape (3,)
Strides sequence for data array
len_pointscnp.npy_intp
Number of points to interpolate
resultarray, float64 shape(N)
The output array. This array should be initialized before you call this function. On exit it will contain the interpolated values from data at points given by points.
Returns:
None
Notes
The output array result is filled in-place.
### nearestneighbor_interpolate
dipy.core.interpolation.nearestneighbor_interpolate()
### trilinear_interp
dipy.core.interpolation.trilinear_interp(data, index, voxel_size)
Interpolates vector from 4D data at 3D point given by index
Interpolates a vector of length T from a 4D volume of shape (I, J, K, T), given point (x, y, z) where (x, y, z) are the coordinates of the point in real units (not yet adjusted for voxel size).
### trilinear_interpolate4d
dipy.core.interpolation.trilinear_interpolate4d()
Tri-linear interpolation along the last dimension of a 4d array
Parameters:
point1d array (3,)
3 doubles representing a 3d point in space. If point has integer values [i, j, k], the result will be the same as data[i, j, k].
data4d array
Data to be interpolated.
out1d array, optional
The output array for the result of the interpolation.
Returns:
out1d array
The result of interpolation.
### as_strided
dipy.core.ndindex.as_strided(x, shape=None, strides=None, subok=False, writeable=True)
Create a view into the array with the given shape and strides.
Warning
This function has to be used with extreme care, see notes.
Parameters:
xndarray
Array to create a new.
shapesequence of int, optional
The shape of the new array. Defaults to x.shape.
stridessequence of int, optional
The strides of the new array. Defaults to x.strides.
subokbool, optional
New in version 1.10.
If True, subclasses are preserved.
writeablebool, optional
New in version 1.12.
If set to False, the returned array will always be readonly. Otherwise it will be writable if the original array was. It is advisable to set this to False if possible (see Notes).
Returns:
viewndarray
broadcast_to
broadcast an array to a given shape.
reshape
reshape an array.
lib.stride_tricks.sliding_window_view
userfriendly and safe function for the creation of sliding window views.
Notes
as_strided creates a view into the array given the exact strides and shape. This means it manipulates the internal data structure of ndarray and, if done incorrectly, the array elements can point to invalid memory and can corrupt results or crash your program. It is advisable to always use the original x.strides when calculating new strides to avoid reliance on a contiguous memory layout.
Furthermore, arrays created with this function often contain self overlapping memory, so that two elements are identical. Vectorized write operations on such arrays will typically be unpredictable. They may even give different results for small, large, or transposed arrays.
Since writing to these arrays has to be tested and done with great care, you may want to use writeable=False to avoid accidental write operations.
For these reasons it is advisable to avoid as_strided when possible.
### ndindex
dipy.core.ndindex.ndindex(shape)
An N-dimensional iterator object to index arrays.
Given the shape of an array, an ndindex instance iterates over the N-dimensional index of the array. At each iteration a tuple of indices is returned; the last dimension is iterated over first.
Parameters:
shapetuple of ints
The dimensions of the array.
Examples
>>> from dipy.core.ndindex import ndindex
>>> shape = (3, 2, 1)
>>> for index in ndindex(shape):
... print(index)
(0, 0, 0)
(0, 1, 0)
(1, 0, 0)
(1, 1, 0)
(2, 0, 0)
(2, 1, 0)
### OneTimeProperty
class dipy.core.onetime.OneTimeProperty(func)
Bases: object
A descriptor to make special properties that become normal attributes.
This is meant to be used mostly by the auto_attr decorator in this module.
__init__(func)
Create a OneTimeProperty instance.
Parameters:
funcmethod
The method that will be called the first time to compute a value.
Afterwards, the method’s name will be a standard attribute holding
the value of this computation.
### ResetMixin
class dipy.core.onetime.ResetMixin
Bases: object
A Mixin class to add a .reset() method to users of OneTimeProperty.
By default, auto attributes once computed, become static. If they happen to depend on other parts of an object and those parts change, their values may now be invalid.
This class offers a .reset() method that users can call explicitly when they know the state of their objects may have changed and they want to ensure that all their special attributes should be invalidated. Once reset() is called, all their auto attributes are reset to their OneTimeProperty descriptors, and their accessor functions will be triggered again.
Warning
If a class has a set of attributes that are OneTimeProperty, but that can be initialized from any one of them, do NOT use this mixin! For instance, UniformTimeSeries can be initialized with only sampling_rate and t0, sampling_interval and time are auto-computed. But if you were to reset() a UniformTimeSeries, it would lose all 4, and there would be then no way to break the circular dependency chains.
If this becomes a problem in practice (for our analyzer objects it isn’t, as they don’t have the above pattern), we can extend reset() to check for a _no_reset set of names in the instance which are meant to be kept protected. But for now this is NOT done, so caveat emptor.
Examples
>>> class A(ResetMixin):
... def __init__(self,x=1.0):
... self.x = x
...
... @auto_attr
... def y(self):
... print('*** y computation executed ***')
... return self.x / 2.0
...
>>> a = A(10)
About to access y twice, the second time no computation is done: >>> a.y * y computation executed * 5.0 >>> a.y 5.0
Changing x >>> a.x = 20
a.y doesn’t change to 10, since it is a static attribute: >>> a.y 5.0
We now reset a, and this will then force all auto attributes to recompute the next time we access them: >>> a.reset()
About to access y twice again after reset(): >>> a.y * y computation executed * 10.0 >>> a.y 10.0
Methods
Reset all OneTimeProperty attributes that may have fired already.
__init__(*args, **kwargs)
reset()
Reset all OneTimeProperty attributes that may have fired already.
### auto_attr
dipy.core.onetime.auto_attr(func)
Decorator to create OneTimeProperty attributes.
Parameters:
funcmethod
The method that will be called the first time to compute a value. Afterwards, the method’s name will be a standard attribute holding the value of this computation.
Examples
>>> class MagicProp(object):
... @auto_attr
... def a(self):
... return 99
...
>>> x = MagicProp()
>>> 'a' in x.__dict__
False
>>> x.a
99
>>> 'a' in x.__dict__
True
### NonNegativeLeastSquares
class dipy.core.optimize.NonNegativeLeastSquares(*args, **kwargs)
A sklearn-like interface to scipy.optimize.nnls
Methods
fit(X, y) Fit the NonNegativeLeastSquares linear model to data predict(X) Predict using the result of the model
__init__(*args, **kwargs)
fit(X, y)
Fit the NonNegativeLeastSquares linear model to data
Parameters:
### Optimizer
class dipy.core.optimize.Optimizer(fun, x0, args=(), method='L-BFGS-B', jac=None, hess=None, hessp=None, bounds=None, constraints=(), tol=None, callback=None, options=None, evolution=False)
Bases: object
Attributes:
evolution
fopt
message
nfev
nit
xopt
Methods
print_summary
__init__(fun, x0, args=(), method='L-BFGS-B', jac=None, hess=None, hessp=None, bounds=None, constraints=(), tol=None, callback=None, options=None, evolution=False)
A class for handling minimization of scalar function of one or more variables.
Parameters:
funcallable
Objective function.
x0ndarray
Initial guess.
argstuple, optional
Extra arguments passed to the objective function and its derivatives (Jacobian, Hessian).
methodstr, optional
Type of solver. Should be one of
• ‘Powell’
• ‘CG’
• ‘BFGS’
• ‘Newton-CG’
• ‘Anneal’
• ‘L-BFGS-B’
• ‘TNC’
• ‘COBYLA’
• ‘SLSQP’
• ‘dogleg’
• ‘trust-ncg’
jacbool or callable, optional
Jacobian of objective function. Only for CG, BFGS, Newton-CG, dogleg, trust-ncg. If jac is a Boolean and is True, fun is assumed to return the value of Jacobian along with the objective function. If False, the Jacobian will be estimated numerically. jac can also be a callable returning the Jacobian of the objective. In this case, it must accept the same arguments as fun.
hess, hesspcallable, optional
Hessian of objective function or Hessian of objective function times an arbitrary vector p. Only for Newton-CG, dogleg, trust-ncg. Only one of hessp or hess needs to be given. If hess is provided, then hessp will be ignored. If neither hess nor hessp is provided, then the hessian product will be approximated using finite differences on jac. hessp must compute the Hessian times an arbitrary vector.
boundssequence, optional
Bounds for variables (only for L-BFGS-B, TNC and SLSQP). (min, max) pairs for each element in x, defining the bounds on that parameter. Use None for one of min or max when there is no bound in that direction.
constraintsdict or sequence of dict, optional
Constraints definition (only for COBYLA and SLSQP). Each constraint is defined in a dictionary with fields:
typestr
Constraint type: ‘eq’ for equality, ‘ineq’ for inequality.
funcallable
The function defining the constraint.
jaccallable, optional
The Jacobian of fun (only for SLSQP).
argssequence, optional
Extra arguments to be passed to the function and Jacobian.
Equality constraint means that the constraint function result is to be zero whereas inequality means that it is to be non-negative. Note that COBYLA only supports inequality constraints.
tolfloat, optional
Tolerance for termination. For detailed control, use solver-specific options.
callbackcallable, optional
Called after each iteration, as callback(xk), where xk is the current parameter vector. Only available using Scipy >= 0.12.
optionsdict, optional
A dictionary of solver options. All methods accept the following generic options:
maxiterint
Maximum number of iterations to perform.
dispbool
Set to True to print convergence messages.
For method-specific options, see show_options(‘minimize’, method).
evolutionbool, optional
save history of x for each iteration. Only available using Scipy >= 0.12.
scipy.optimize.minimize
property evolution
property fopt
property message
property nfev
property nit
print_summary()
property xopt
### PositiveDefiniteLeastSquares
class dipy.core.optimize.PositiveDefiniteLeastSquares(m, A=None, L=None)
Bases: object
Methods
solve(design_matrix, measurements[, check]) Solve CVXPY problem
__init__(m, A=None, L=None)
Regularized least squares with linear matrix inequality constraints
Generate a CVXPY representation of a regularized least squares optimization problem subject to linear matrix inequality constraints.
Parameters:
mint
Positive int indicating the number of regressors.
Aarray (t = m + k + 1, p, p) (optional)
Constraint matrices $$A$$.
Larray (m, m) (optional)
Regularization matrix $$L$$. Default: None.
Notes
The basic problem is to solve for $$h$$ the minimization of
$$c=\|X h - y\|^2 + \|L h\|^2$$,
where $$X$$ is an (m, m) upper triangular design matrix and $$y$$ is a set of m measurements, subject to the constraint that
$$M=A_0+\sum_{i=0}^{m-1} h_i A_{i+1}+\sum_{j=0}^{k-1} s_j A_{m+j+1}>0$$,
where $$s_j$$ are slack variables and where the inequality sign denotes positive definiteness of the matrix $$M$$. The sparsity pattern and size of $$X$$ and $$y$$ are fixed, because every design matrix and set of measurements can be reduced to an equivalent (minimal) formulation of this type.
This formulation is used here mainly to enforce polynomial sum-of-squares constraints on various models, as described in [1].
References
[1]
Dela Haije et al. “Enforcing necessary non-negativity constraints for common diffusion MRI models using sum of squares programming”. NeuroImage 209, 2020, 116405.
solve(design_matrix, measurements, check=False, **kwargs)
Solve CVXPY problem
Solve a CVXPY problem instance for a given design matrix and a given set of observations, and return the optimum.
Parameters:
design_matrixarray (n, m)
Design matrix.
measurementsarray (n)
Measurements.
checkboolean (optional)
If True check whether the unconstrained optimization solution already satisfies the constraints, before running the constrained optimization. This adds overhead, but can avoid unnecessary constrained optimization calls. Default: False
kwargskeyword arguments
Arguments passed to the CVXPY solve method.
Returns:
harray (m)
Estimated optimum for problem variables $$h$$.
### SKLearnLinearSolver
class dipy.core.optimize.SKLearnLinearSolver(*args, **kwargs)
Bases: object
Provide a sklearn-like uniform interface to algorithms that solve problems of the form: $$y = Ax$$ for $$x$$
Sub-classes of SKLearnLinearSolver should provide a ‘fit’ method that have the following signature: SKLearnLinearSolver.fit(X, y), which would set an attribute SKLearnLinearSolver.coef_, with the shape (X.shape[1],), such that an estimate of y can be calculated as: y_hat = np.dot(X, SKLearnLinearSolver.coef_.T)
Methods
fit(X, y) Implement for all derived classes Predict using the result of the model
__init__(*args, **kwargs)
abstract fit(X, y)
Implement for all derived classes
predict(X)
Predict using the result of the model
Parameters:
Xarray-like (n_samples, n_features)
Samples.
Returns:
Carray, shape = (n_samples,)
Predicted values.
### Version
class dipy.core.optimize.Version(version: str)
Bases: _BaseVersion
This class abstracts handling of a project’s versions.
A Version instance is comparison aware and can be compared and sorted using the standard Python interfaces.
>>> v1 = Version("1.0a5")
>>> v2 = Version("1.0")
>>> v1
<Version('1.0a5')>
>>> v2
<Version('1.0')>
>>> v1 < v2
True
>>> v1 == v2
False
>>> v1 > v2
False
>>> v1 >= v2
False
>>> v1 <= v2
True
Attributes:
base_version
The “base version” of the version.
dev
The development number of the version.
epoch
The epoch of the version.
is_devrelease
Whether this version is a development release.
is_postrelease
Whether this version is a post-release.
is_prerelease
Whether this version is a pre-release.
local
The local version segment of the version.
major
The first item of release or 0 if unavailable.
micro
The third item of release or 0 if unavailable.
minor
The second item of release or 0 if unavailable.
post
The post-release number of the version.
pre
The pre-release segment of the version.
public
The public portion of the version.
release
The components of the “release” segment of the version.
__init__(version: str) None
Initialize a Version object.
Parameters:
version – The string representation of a version which will be parsed and normalized before use.
Raises:
InvalidVersion – If the version does not conform to PEP 440 in any way then this exception will be raised.
property base_version: str
The “base version” of the version.
>>> Version("1.2.3").base_version
'1.2.3'
>>> Version("1.2.3+abc").base_version
'1.2.3'
>>> Version("1!1.2.3+abc.dev1").base_version
'1!1.2.3'
The “base version” is the public version of the project without any pre or post release markers.
property dev: int | None
The development number of the version.
>>> print(Version("1.2.3").dev)
None
>>> Version("1.2.3.dev1").dev
1
property epoch: int
The epoch of the version.
>>> Version("2.0.0").epoch
0
>>> Version("1!2.0.0").epoch
1
property is_devrelease: bool
Whether this version is a development release.
>>> Version("1.2.3").is_devrelease
False
>>> Version("1.2.3.dev1").is_devrelease
True
property is_postrelease: bool
Whether this version is a post-release.
>>> Version("1.2.3").is_postrelease
False
>>> Version("1.2.3.post1").is_postrelease
True
property is_prerelease: bool
Whether this version is a pre-release.
>>> Version("1.2.3").is_prerelease
False
>>> Version("1.2.3a1").is_prerelease
True
>>> Version("1.2.3b1").is_prerelease
True
>>> Version("1.2.3rc1").is_prerelease
True
>>> Version("1.2.3dev1").is_prerelease
True
property local: str | None
The local version segment of the version.
>>> print(Version("1.2.3").local)
None
>>> Version("1.2.3+abc").local
'abc'
property major: int
The first item of release or 0 if unavailable.
>>> Version("1.2.3").major
1
property micro: int
The third item of release or 0 if unavailable.
>>> Version("1.2.3").micro
3
>>> Version("1").micro
0
property minor: int
The second item of release or 0 if unavailable.
>>> Version("1.2.3").minor
2
>>> Version("1").minor
0
property post: int | None
The post-release number of the version.
>>> print(Version("1.2.3").post)
None
>>> Version("1.2.3.post1").post
1
property pre: Tuple[str, int] | None
The pre-release segment of the version.
>>> print(Version("1.2.3").pre)
None
>>> Version("1.2.3a1").pre
('a', 1)
>>> Version("1.2.3b1").pre
('b', 1)
>>> Version("1.2.3rc1").pre
('rc', 1)
property public: str
The public portion of the version.
>>> Version("1.2.3").public
'1.2.3'
>>> Version("1.2.3+abc").public
'1.2.3'
>>> Version("1.2.3+abc.dev1").public
'1.2.3'
property release: Tuple[int, ...]
The components of the “release” segment of the version.
>>> Version("1.2.3").release
(1, 2, 3)
>>> Version("2.0.0").release
(2, 0, 0)
>>> Version("1!2.0.0.post0").release
(2, 0, 0)
Includes trailing zeroes but not the epoch or any pre-release / development / post-release suffixes.
### minimize
dipy.core.optimize.minimize(fun, x0, args=(), method=None, jac=None, hess=None, hessp=None, bounds=None, constraints=(), tol=None, callback=None, options=None)
Minimization of scalar function of one or more variables.
Parameters:
funcallable
The objective function to be minimized.
fun(x, *args) -> float
where x is a 1-D array with shape (n,) and args is a tuple of the fixed parameters needed to completely specify the function.
x0ndarray, shape (n,)
Initial guess. Array of real elements of size (n,), where n is the number of independent variables.
argstuple, optional
Extra arguments passed to the objective function and its derivatives (fun, jac and hess functions).
methodstr or callable, optional
Type of solver. Should be one of
• ‘Powell’ (see here)
• ‘CG’ (see here)
• ‘BFGS’ (see here)
• ‘Newton-CG’ (see here)
• ‘L-BFGS-B’ (see here)
• ‘TNC’ (see here)
• ‘COBYLA’ (see here)
• ‘SLSQP’ (see here)
• ‘trust-constr’(see here)
• ‘dogleg’ (see here)
• ‘trust-ncg’ (see here)
• ‘trust-exact’ (see here)
• ‘trust-krylov’ (see here)
• custom - a callable object, see below for description.
If not given, chosen to be one of BFGS, L-BFGS-B, SLSQP, depending on whether or not the problem has constraints or bounds.
jac{callable, ‘2-point’, ‘3-point’, ‘cs’, bool}, optional
Method for computing the gradient vector. Only for CG, BFGS, Newton-CG, L-BFGS-B, TNC, SLSQP, dogleg, trust-ncg, trust-krylov, trust-exact and trust-constr. If it is a callable, it should be a function that returns the gradient vector:
jac(x, *args) -> array_like, shape (n,)
where x is an array with shape (n,) and args is a tuple with the fixed parameters. If jac is a Boolean and is True, fun is assumed to return a tuple (f, g) containing the objective function and the gradient. Methods ‘Newton-CG’, ‘trust-ncg’, ‘dogleg’, ‘trust-exact’, and ‘trust-krylov’ require that either a callable be supplied, or that fun return the objective and gradient. If None or False, the gradient will be estimated using 2-point finite difference estimation with an absolute step size. Alternatively, the keywords {‘2-point’, ‘3-point’, ‘cs’} can be used to select a finite difference scheme for numerical estimation of the gradient with a relative step size. These finite difference schemes obey any specified bounds.
hess{callable, ‘2-point’, ‘3-point’, ‘cs’, HessianUpdateStrategy}, optional
Method for computing the Hessian matrix. Only for Newton-CG, dogleg, trust-ncg, trust-krylov, trust-exact and trust-constr. If it is callable, it should return the Hessian matrix:
hess(x, *args) -> {LinearOperator, spmatrix, array}, (n, n)
where x is a (n,) ndarray and args is a tuple with the fixed parameters. The keywords {‘2-point’, ‘3-point’, ‘cs’} can also be used to select a finite difference scheme for numerical estimation of the hessian. Alternatively, objects implementing the HessianUpdateStrategy interface can be used to approximate the Hessian. Available quasi-Newton methods implementing this interface are:
• BFGS;
• SR1.
Not all of the options are available for each of the methods; for availability refer to the notes.
hesspcallable, optional
Hessian of objective function times an arbitrary vector p. Only for Newton-CG, trust-ncg, trust-krylov, trust-constr. Only one of hessp or hess needs to be given. If hess is provided, then hessp will be ignored. hessp must compute the Hessian times an arbitrary vector:
hessp(x, p, *args) -> ndarray shape (n,)
where x is a (n,) ndarray, p is an arbitrary vector with dimension (n,) and args is a tuple with the fixed parameters.
boundssequence or Bounds, optional
Bounds on variables for Nelder-Mead, L-BFGS-B, TNC, SLSQP, Powell, and trust-constr methods. There are two ways to specify the bounds:
1. Instance of Bounds class.
2. Sequence of (min, max) pairs for each element in x. None is used to specify no bound.
constraints{Constraint, dict} or List of {Constraint, dict}, optional
Constraints definition. Only for COBYLA, SLSQP and trust-constr.
Constraints for ‘trust-constr’ are defined as a single object or a list of objects specifying constraints to the optimization problem. Available constraints are:
• LinearConstraint
• NonlinearConstraint
Constraints for COBYLA, SLSQP are defined as a list of dictionaries. Each dictionary with fields:
typestr
Constraint type: ‘eq’ for equality, ‘ineq’ for inequality.
funcallable
The function defining the constraint.
jaccallable, optional
The Jacobian of fun (only for SLSQP).
argssequence, optional
Extra arguments to be passed to the function and Jacobian.
Equality constraint means that the constraint function result is to be zero whereas inequality means that it is to be non-negative. Note that COBYLA only supports inequality constraints.
tolfloat, optional
Tolerance for termination. When tol is specified, the selected minimization algorithm sets some relevant solver-specific tolerance(s) equal to tol. For detailed control, use solver-specific options.
optionsdict, optional
A dictionary of solver options. All methods except TNC accept the following generic options:
maxiterint
Maximum number of iterations to perform. Depending on the method each iteration may use several function evaluations.
For TNC use maxfun instead of maxiter.
dispbool
Set to True to print convergence messages.
For method-specific options, see show_options().
callbackcallable, optional
Called after each iteration. For ‘trust-constr’ it is a callable with the signature:
callback(xk, OptimizeResult state) -> bool
where xk is the current parameter vector. and state is an OptimizeResult object, with the same fields as the ones from the return. If callback returns True the algorithm execution is terminated. For all the other methods, the signature is:
callback(xk)
where xk is the current parameter vector.
Returns:
resOptimizeResult
The optimization result represented as a OptimizeResult object. Important attributes are: x the solution array, success a Boolean flag indicating if the optimizer exited successfully and message which describes the cause of the termination. See OptimizeResult for a description of other attributes.
minimize_scalar
Interface to minimization algorithms for scalar univariate functions
show_options
Additional options accepted by the solvers
Notes
This section describes the available solvers that can be selected by the ‘method’ parameter. The default method is BFGS.
Unconstrained minimization
Method CG uses a nonlinear conjugate gradient algorithm by Polak and Ribiere, a variant of the Fletcher-Reeves method described in [5] pp.120-122. Only the first derivatives are used.
Method BFGS uses the quasi-Newton method of Broyden, Fletcher, Goldfarb, and Shanno (BFGS) [5] pp. 136. It uses the first derivatives only. BFGS has proven good performance even for non-smooth optimizations. This method also returns an approximation of the Hessian inverse, stored as hess_inv in the OptimizeResult object.
Method Newton-CG uses a Newton-CG algorithm [5] pp. 168 (also known as the truncated Newton method). It uses a CG method to the compute the search direction. See also TNC method for a box-constrained minimization with a similar algorithm. Suitable for large-scale problems.
Method dogleg uses the dog-leg trust-region algorithm [5] for unconstrained minimization. This algorithm requires the gradient and Hessian; furthermore the Hessian is required to be positive definite.
Method trust-ncg uses the Newton conjugate gradient trust-region algorithm [5] for unconstrained minimization. This algorithm requires the gradient and either the Hessian or a function that computes the product of the Hessian with a given vector. Suitable for large-scale problems.
Method trust-krylov uses the Newton GLTR trust-region algorithm [14], [15] for unconstrained minimization. This algorithm requires the gradient and either the Hessian or a function that computes the product of the Hessian with a given vector. Suitable for large-scale problems. On indefinite problems it requires usually less iterations than the trust-ncg method and is recommended for medium and large-scale problems.
Method trust-exact is a trust-region method for unconstrained minimization in which quadratic subproblems are solved almost exactly [13]. This algorithm requires the gradient and the Hessian (which is not required to be positive definite). It is, in many situations, the Newton method to converge in fewer iterations and the most recommended for small and medium-size problems.
Bound-Constrained minimization
Method Nelder-Mead uses the Simplex algorithm [1], [2]. This algorithm is robust in many applications. However, if numerical computation of derivative can be trusted, other algorithms using the first and/or second derivatives information might be preferred for their better performance in general.
Method L-BFGS-B uses the L-BFGS-B algorithm [6], [7] for bound constrained minimization.
Method Powell is a modification of Powell’s method [3], [4] which is a conjugate direction method. It performs sequential one-dimensional minimizations along each vector of the directions set (direc field in options and info), which is updated at each iteration of the main minimization loop. The function need not be differentiable, and no derivatives are taken. If bounds are not provided, then an unbounded line search will be used. If bounds are provided and the initial guess is within the bounds, then every function evaluation throughout the minimization procedure will be within the bounds. If bounds are provided, the initial guess is outside the bounds, and direc is full rank (default has full rank), then some function evaluations during the first iteration may be outside the bounds, but every function evaluation after the first iteration will be within the bounds. If direc is not full rank, then some parameters may not be optimized and the solution is not guaranteed to be within the bounds.
Method TNC uses a truncated Newton algorithm [5], [8] to minimize a function with variables subject to bounds. This algorithm uses gradient information; it is also called Newton Conjugate-Gradient. It differs from the Newton-CG method described above as it wraps a C implementation and allows each variable to be given upper and lower bounds.
Constrained Minimization
Method COBYLA uses the Constrained Optimization BY Linear Approximation (COBYLA) method [9], [10], [11]. The algorithm is based on linear approximations to the objective function and each constraint. The method wraps a FORTRAN implementation of the algorithm. The constraints functions ‘fun’ may return either a single number or an array or list of numbers.
Method SLSQP uses Sequential Least SQuares Programming to minimize a function of several variables with any combination of bounds, equality and inequality constraints. The method wraps the SLSQP Optimization subroutine originally implemented by Dieter Kraft [12]. Note that the wrapper handles infinite values in bounds by converting them into large floating values.
Method trust-constr is a trust-region algorithm for constrained optimization. It swiches between two implementations depending on the problem definition. It is the most versatile constrained minimization algorithm implemented in SciPy and the most appropriate for large-scale problems. For equality constrained problems it is an implementation of Byrd-Omojokun Trust-Region SQP method described in [17] and in [5], p. 549. When inequality constraints are imposed as well, it swiches to the trust-region interior point method described in [16]. This interior point algorithm, in turn, solves inequality constraints by introducing slack variables and solving a sequence of equality-constrained barrier problems for progressively smaller values of the barrier parameter. The previously described equality constrained SQP method is used to solve the subproblems with increasing levels of accuracy as the iterate gets closer to a solution.
Finite-Difference Options
For Method trust-constr the gradient and the Hessian may be approximated using three finite-difference schemes: {‘2-point’, ‘3-point’, ‘cs’}. The scheme ‘cs’ is, potentially, the most accurate but it requires the function to correctly handle complex inputs and to be differentiable in the complex plane. The scheme ‘3-point’ is more accurate than ‘2-point’ but requires twice as many operations. If the gradient is estimated via finite-differences the Hessian must be estimated using one of the quasi-Newton strategies.
Method specific options for the hess keyword
method/Hess
None
callable
‘2-point/’3-point’/’cs’
HUS
Newton-CG
x
(n, n) LO
x
x
dogleg
(n, n)
trust-ncg
(n, n)
x
x
trust-krylov
(n, n)
x
x
trust-exact
(n, n)
trust-constr
x
(n, n) LO sp
x
x
Custom minimizers
It may be useful to pass a custom minimization method, for example when using a frontend to this method such as scipy.optimize.basinhopping or a different library. You can simply pass a callable as the method parameter.
The callable is called as method(fun, x0, args, **kwargs, **options) where kwargs corresponds to any other parameters passed to minimize (such as callback, hess, etc.), except the options dict, which has its contents also passed as method parameters pair by pair. Also, if jac has been passed as a bool type, jac and fun are mangled so that fun returns just the function values and jac is converted to a function returning the Jacobian. The method shall return an OptimizeResult object.
The provided method callable must be able to accept (and possibly ignore) arbitrary parameters; the set of parameters accepted by minimize may expand in future versions and then these parameters will be passed to the method. You can find an example in the scipy.optimize tutorial.
References
[1]
Nelder, J A, and R Mead. 1965. A Simplex Method for Function Minimization. The Computer Journal 7: 308-13.
[2]
Wright M H. 1996. Direct search methods: Once scorned, now respectable, in Numerical Analysis 1995: Proceedings of the 1995 Dundee Biennial Conference in Numerical Analysis (Eds. D F Griffiths and G A Watson). Addison Wesley Longman, Harlow, UK. 191-208.
[3]
Powell, M J D. 1964. An efficient method for finding the minimum of a function of several variables without calculating derivatives. The Computer Journal 7: 155-162.
[4]
Press W, S A Teukolsky, W T Vetterling and B P Flannery. Numerical Recipes (any edition), Cambridge University Press.
[5] (1,2,3,4,5,6,7,8)
Nocedal, J, and S J Wright. 2006. Numerical Optimization. Springer New York.
[6]
Byrd, R H and P Lu and J. Nocedal. 1995. A Limited Memory Algorithm for Bound Constrained Optimization. SIAM Journal on Scientific and Statistical Computing 16 (5): 1190-1208.
[7]
Zhu, C and R H Byrd and J Nocedal. 1997. L-BFGS-B: Algorithm 778: L-BFGS-B, FORTRAN routines for large scale bound constrained optimization. ACM Transactions on Mathematical Software 23 (4): 550-560.
[8]
Nash, S G. Newton-Type Minimization Via the Lanczos Method. 1984. SIAM Journal of Numerical Analysis 21: 770-778.
[9]
Powell, M J D. A direct search optimization method that models the objective and constraint functions by linear interpolation. 1994. Advances in Optimization and Numerical Analysis, eds. S. Gomez and J-P Hennart, Kluwer Academic (Dordrecht), 51-67.
[10]
Powell M J D. Direct search algorithms for optimization calculations. 1998. Acta Numerica 7: 287-336.
[11]
Powell M J D. A view of algorithms for optimization without derivatives. 2007.Cambridge University Technical Report DAMTP 2007/NA03
[12]
Kraft, D. A software package for sequential quadratic programming. 1988. Tech. Rep. DFVLR-FB 88-28, DLR German Aerospace Center – Institute for Flight Mechanics, Koln, Germany.
[13]
Conn, A. R., Gould, N. I., and Toint, P. L. Trust region methods. 2000. Siam. pp. 169-200.
[14]
F. Lenders, C. Kirches, A. Potschka: “trlib: A vector-free implementation of the GLTR method for iterative solution of the trust region problem”, :arxiv:1611.04718
[15]
N. Gould, S. Lucidi, M. Roma, P. Toint: “Solving the Trust-Region Subproblem using the Lanczos Method”, SIAM J. Optim., 9(2), 504–525, (1999).
[16]
Byrd, Richard H., Mary E. Hribar, and Jorge Nocedal. 1999. An interior point algorithm for large-scale nonlinear programming. SIAM Journal on Optimization 9.4: 877-900.
[17]
Lalee, Marucha, Jorge Nocedal, and Todd Plantega. 1998. On the implementation of an algorithm for large-scale equality constrained optimization. SIAM Journal on Optimization 8.3: 682-706.
Examples
Let us consider the problem of minimizing the Rosenbrock function. This function (and its respective derivatives) is implemented in rosen (resp. rosen_der, rosen_hess) in the scipy.optimize.
>>> from scipy.optimize import minimize, rosen, rosen_der
A simple application of the Nelder-Mead method is:
>>> x0 = [1.3, 0.7, 0.8, 1.9, 1.2]
>>> res = minimize(rosen, x0, method='Nelder-Mead', tol=1e-6)
>>> res.x
array([ 1., 1., 1., 1., 1.])
Now using the BFGS algorithm, using the first derivative and a few options:
>>> res = minimize(rosen, x0, method='BFGS', jac=rosen_der,
... options={'gtol': 1e-6, 'disp': True})
Optimization terminated successfully.
Current function value: 0.000000
Iterations: 26
Function evaluations: 31
>>> res.x
array([ 1., 1., 1., 1., 1.])
>>> print(res.message)
Optimization terminated successfully.
>>> res.hess_inv
array([[ 0.00749589, 0.01255155, 0.02396251, 0.04750988, 0.09495377], # may vary
[ 0.01255155, 0.02510441, 0.04794055, 0.09502834, 0.18996269],
[ 0.02396251, 0.04794055, 0.09631614, 0.19092151, 0.38165151],
[ 0.04750988, 0.09502834, 0.19092151, 0.38341252, 0.7664427 ],
[ 0.09495377, 0.18996269, 0.38165151, 0.7664427, 1.53713523]])
Next, consider a minimization problem with several constraints (namely Example 16.4 from [5]). The objective function is:
>>> fun = lambda x: (x[0] - 1)**2 + (x[1] - 2.5)**2
There are three constraints defined as:
>>> cons = ({'type': 'ineq', 'fun': lambda x: x[0] - 2 * x[1] + 2},
... {'type': 'ineq', 'fun': lambda x: -x[0] - 2 * x[1] + 6},
... {'type': 'ineq', 'fun': lambda x: -x[0] + 2 * x[1] + 2})
And variables must be positive, hence the following bounds:
>>> bnds = ((0, None), (0, None))
The optimization problem is solved using the SLSQP method as:
>>> res = minimize(fun, (2, 0), method='SLSQP', bounds=bnds,
... constraints=cons)
It should converge to the theoretical solution (1.4 ,1.7).
### optional_package
dipy.core.optimize.optional_package(name, trip_msg=None)
Return package-like thing and module setup for package name
Parameters:
namestr
package name
trip_msgNone or str
message to give when someone tries to use the return package, but we could not import it, and have returned a TripWire object instead. Default message if None.
Returns:
pkg_likemodule or TripWire instance
If we can import the package, return it. Otherwise return an object raising an error when accessed
have_pkgbool
True if import for package was successful, false otherwise
module_setupfunction
callable usually set as setup_module in calling namespace, to allow skipping tests.
Examples
Typical use would be something like this at the top of a module using an optional package:
>>> from dipy.utils.optpkg import optional_package
>>> pkg, have_pkg, setup_module = optional_package('not_a_package')
Of course in this case the package doesn’t exist, and so, in the module:
>>> have_pkg
False
and
>>> pkg.some_function()
Traceback (most recent call last):
...
TripWireError: We need package not_a_package for these functions, but
import not_a_package raised an ImportError
If the module does exist - we get the module
>>> pkg, _, _ = optional_package('os')
>>> hasattr(pkg, 'path')
True
Or a submodule if that’s what we asked for
>>> subpkg, _, _ = optional_package('os.path')
>>> hasattr(subpkg, 'dirname')
True
### sparse_nnls
dipy.core.optimize.sparse_nnls(y, X, momentum=1, step_size=0.01, non_neg=True, check_error_iter=10, max_error_checks=10, converge_on_sse=0.99)
Solve y=Xh for h, using gradient descent, with X a sparse matrix.
Parameters:
y1-d array of shape (N)
The data. Needs to be dense.
Xndarray. May be either sparse or dense. Shape (N, M)
The regressors
momentumfloat, optional (default: 1).
step_sizefloat, optional (default: 0.01).
The increment of parameter update in each iteration
non_negBoolean, optional (default: True)
Whether to enforce non-negativity of the solution.
check_error_iterint (default:10)
How many rounds to run between error evaluation for convergence-checking.
max_error_checksint (default: 10)
Don’t check errors more than this number of times if no improvement in r-squared is seen.
converge_on_ssefloat (default: 0.99)
a percentage improvement in SSE that is required each time to say that things are still going well.
Returns:
h_bestThe best estimate of the parameters.
### spdot
dipy.core.optimize.spdot(A, B)
The same as np.dot(A, B), except it works even if A or B or both are sparse matrices.
Parameters:
A, Barrays of shape (m, n), (n, k)
Returns:
The matrix product AB. If both A and B are sparse, the result will be a
sparse matrix. Otherwise, a dense result is returned
See discussion here:
http://mail.scipy.org/pipermail/scipy-user/2010-November/027700.html
### Profiler
class dipy.core.profile.Profiler(call=None, *args)
Bases: object
Profile python/cython files or functions
If you are profiling cython code you need to add # cython: profile=True on the top of your .pyx file
and for the functions that you do not want to profile you can use this decorator in your cython files
@cython.profile(False)
Parameters:
callerfile or function call
argsfunction arguments
References
Examples
from dipy.core.profile import Profiler import numpy as np p=Profiler(np.sum,np.random.rand(1000000,3)) fname=’test.py’ p=Profiler(fname) p.print_stats(10) p.print_stats(‘det’)
Attributes:
statsfunction, stats.print_stats(10) will prin the 10 slower functions
Methods
Print stats for profiling
__init__(call=None, *args)
print_stats(N=10)
Print stats for profiling
You can use it in all different ways developed in pstats for example print_stats(10) will give you the 10 slowest calls or print_stats(‘function_name’) will give you the stats for all the calls with name ‘function_name’
Parameters:
Nstats.print_stats argument
### optional_package
dipy.core.profile.optional_package(name, trip_msg=None)
Return package-like thing and module setup for package name
Parameters:
namestr
package name
trip_msgNone or str
message to give when someone tries to use the return package, but we could not import it, and have returned a TripWire object instead. Default message if None.
Returns:
pkg_likemodule or TripWire instance
If we can import the package, return it. Otherwise return an object raising an error when accessed
have_pkgbool
True if import for package was successful, false otherwise
module_setupfunction
callable usually set as setup_module in calling namespace, to allow skipping tests.
Examples
Typical use would be something like this at the top of a module using an optional package:
>>> from dipy.utils.optpkg import optional_package
>>> pkg, have_pkg, setup_module = optional_package('not_a_package')
Of course in this case the package doesn’t exist, and so, in the module:
>>> have_pkg
False
and
>>> pkg.some_function()
Traceback (most recent call last):
...
TripWireError: We need package not_a_package for these functions, but
import not_a_package raised an ImportError
If the module does exist - we get the module
>>> pkg, _, _ = optional_package('os')
>>> hasattr(pkg, 'path')
True
Or a submodule if that’s what we asked for
>>> subpkg, _, _ = optional_package('os.path')
>>> hasattr(subpkg, 'dirname')
True
### LEcuyer
dipy.core.rng.LEcuyer(s1=100001, s2=200002)
Return a LEcuyer random number generator.
Generate uniformly distributed random numbers using the 32-bit generator from figure 3 of:
L’Ecuyer, P. Efficient and portable combined random number generators, C.A.C.M., vol. 31, 742-749 & 774-?, June 1988.
The cycle length is claimed to be 2.30584E+18
Parameters:
s1: int
First seed value. Should not be null. (default 100001)
s2: int
Second seed value. Should not be null. (default 200002)
Returns:
r_numberfloat
pseudo-random number uniformly distributed between [0-1]
Examples
>>> from dipy.core import rng
>>> N = 1000
>>> a = [rng.LEcuyer() for i in range(N)]
### WichmannHill1982
dipy.core.rng.WichmannHill1982(ix=100001, iy=200002, iz=300003)
Algorithm AS 183 Appl. Statist. (1982) vol.31, no.2.
Returns a pseudo-random number rectangularly distributed between 0 and 1. The cycle length is 6.95E+12 (See page 123 of Applied Statistics (1984) vol.33), not as claimed in the original article.
ix, iy and iz should be set to integer values between 1 and 30000 before the first entry.
Integer arithmetic up to 5212632 is required.
Parameters:
ix: int
First seed value. Should not be null. (default 100001)
iy: int
Second seed value. Should not be null. (default 200002)
iz: int
Third seed value. Should not be null. (default 300003)
Returns:
r_numberfloat
pseudo-random number uniformly distributed between [0-1]
Examples
>>> from dipy.core import rng
>>> N = 1000
>>> a = [rng.WichmannHill1982() for i in range(N)]
### WichmannHill2006
dipy.core.rng.WichmannHill2006(ix=100001, iy=200002, iz=300003, it=400004)
Wichmann Hill (2006) random number generator.
B.A. Wichmann, I.D. Hill, Generating good pseudo-random numbers, Computational Statistics & Data Analysis, Volume 51, Issue 3, 1 December 2006, Pages 1614-1622, ISSN 0167-9473, DOI: 10.1016/j.csda.2006.05.019. (http://www.sciencedirect.com/science/article/B6V8V-4K7F86W-2/2/a3a33291b8264e4c882a8f21b6e43351) for advice on generating many sequences for use together, and on alternative algorithms and codes
Parameters:
ix: int
First seed value. Should not be null. (default 100001)
iy: int
Second seed value. Should not be null. (default 200002)
iz: int
Third seed value. Should not be null. (default 300003)
it: int
Fourth seed value. Should not be null. (default 400004)
Returns:
r_numberfloat
pseudo-random number uniformly distributed between [0-1]
Examples
>>> from dipy.core import rng
>>> N = 1000
>>> a = [rng.WichmannHill2006() for i in range(N)]
### architecture
Queries the given executable (defaults to the Python interpreter binary) for various architecture information.
Returns a tuple (bits, linkage) which contains information about the bit architecture and the linkage format used for the executable. Both values are returned as strings.
Values that cannot be determined are returned as given by the parameter presets. If bits is given as ‘’, the sizeof(pointer) (or sizeof(long) on Python version < 1.5.2) is used as indicator for the supported pointer size.
The function relies on the system’s “file” command to do the actual work. This is available on most if not all Unix platforms. On some non-Unix platforms where the “file” command does not exist and the executable is set to the Python interpreter binary defaults from _default_architecture are used.
### floor
dipy.core.rng.floor(x, /)
Return the floor of x as an Integral.
This is the largest integer <= x.
### HemiSphere
class dipy.core.sphere.HemiSphere(x=None, y=None, z=None, theta=None, phi=None, xyz=None, faces=None, edges=None, tol=1e-05)
Bases: Sphere
Points on the unit sphere.
A HemiSphere is similar to a Sphere but it takes antipodal symmetry into account. Antipodal symmetry means that point v on a HemiSphere is the same as the point -v. Duplicate points are discarded when constructing a HemiSphere (including antipodal duplicates). edges and faces are remapped to the remaining points as closely as possible.
The HemiSphere can be constructed using one of three conventions:
HemiSphere(x, y, z)
HemiSphere(xyz=xyz)
HemiSphere(theta=theta, phi=phi)
Parameters:
x, y, z1-D array_like
Vertices as x-y-z coordinates.
theta, phi1-D array_like
Vertices as spherical coordinates. Theta and phi are the inclination and azimuth angles respectively.
xyz(N, 3) ndarray
Vertices as x-y-z coordinates.
faces(N, 3) ndarray
Indices into vertices that form triangular faces. If unspecified, the faces are computed using a Delaunay triangulation.
edges(N, 2) ndarray
Edges between vertices. If unspecified, the edges are derived from the faces.
tolfloat
Angle in degrees. Vertices that are less than tol degrees apart are treated as duplicates.
Attributes:
x
y
z
Methods
Find the index of the vertex in the Sphere closest to the input vector, taking into account antipodal symmetry from_sphere(sphere[, tol]) Create instance from a Sphere Create a full Sphere from a HemiSphere subdivide([n]) Create a more subdivided HemiSphere
edges faces vertices
__init__(x=None, y=None, z=None, theta=None, phi=None, xyz=None, faces=None, edges=None, tol=1e-05)
Create a HemiSphere from points
faces()
find_closest(xyz)
Find the index of the vertex in the Sphere closest to the input vector, taking into account antipodal symmetry
Parameters:
xyzarray-like, 3 elements
A unit vector
Returns:
idxint
The index into the Sphere.vertices array that gives the closest vertex (in angle).
classmethod from_sphere(sphere, tol=1e-05)
Create instance from a Sphere
mirror()
Create a full Sphere from a HemiSphere
subdivide(n=1)
Create a more subdivided HemiSphere
See Sphere.subdivide for full documentation.
### Sphere
class dipy.core.sphere.Sphere(x=None, y=None, z=None, theta=None, phi=None, xyz=None, faces=None, edges=None)
Bases: object
Points on the unit sphere.
The sphere can be constructed using one of three conventions:
Sphere(x, y, z)
Sphere(xyz=xyz)
Sphere(theta=theta, phi=phi)
Parameters:
x, y, z1-D array_like
Vertices as x-y-z coordinates.
theta, phi1-D array_like
Vertices as spherical coordinates. Theta and phi are the inclination and azimuth angles respectively.
xyz(N, 3) ndarray
Vertices as x-y-z coordinates.
faces(N, 3) ndarray
Indices into vertices that form triangular faces. If unspecified, the faces are computed using a Delaunay triangulation.
edges(N, 2) ndarray
Edges between vertices. If unspecified, the edges are derived from the faces.
Attributes:
x
y
z
Methods
Find the index of the vertex in the Sphere closest to the input vector subdivide([n]) Subdivides each face of the sphere into four new faces.
edges faces vertices
__init__(x=None, y=None, z=None, theta=None, phi=None, xyz=None, faces=None, edges=None)
edges()
faces()
find_closest(xyz)
Find the index of the vertex in the Sphere closest to the input vector
Parameters:
xyzarray-like, 3 elements
A unit vector
Returns:
idxint
The index into the Sphere.vertices array that gives the closest vertex (in angle).
subdivide(n=1)
Subdivides each face of the sphere into four new faces.
New vertices are created at a, b, and c. Then each face [x, y, z] is divided into faces [x, a, c], [y, a, b], [z, b, c], and [a, b, c].
y
/\
/ \
a/____\b
/\ /\
/ \ / \
/____\/____\
x c z
Parameters:
nint, optional
The number of subdivisions to preform.
Returns:
new_sphereSphere
The subdivided sphere.
vertices()
property x
property y
property z
### auto_attr
dipy.core.sphere.auto_attr(func)
Decorator to create OneTimeProperty attributes.
Parameters:
funcmethod
The method that will be called the first time to compute a value. Afterwards, the method’s name will be a standard attribute holding the value of this computation.
Examples
>>> class MagicProp(object):
... @auto_attr
... def a(self):
... return 99
...
>>> x = MagicProp()
>>> 'a' in x.__dict__
False
>>> x.a
99
>>> 'a' in x.__dict__
True
### cart2sphere
dipy.core.sphere.cart2sphere(x, y, z)
Return angles for Cartesian 3D coordinates x, y, and z
See doc for sphere2cart for angle conventions and derivation of the formulae.
$$0\le\theta\mathrm{(theta)}\le\pi$$ and $$-\pi\le\phi\mathrm{(phi)}\le\pi$$
Parameters:
xarray_like
x coordinate in Cartesian space
yarray_like
y coordinate in Cartesian space
zarray_like
z coordinate
Returns:
rarray
thetaarray
inclination (polar) angle
phiarray
azimuth angle
### disperse_charges
dipy.core.sphere.disperse_charges(hemi, iters, const=0.2)
Models electrostatic repulsion on the unit sphere
Places charges on a sphere and simulates the repulsive forces felt by each one. Allows the charges to move for some number of iterations and returns their final location as well as the total potential of the system at each step.
Parameters:
hemiHemiSphere
Points on a unit sphere.
itersint
Number of iterations to run.
constfloat
Using a smaller const could provide a more accurate result, but will need more iterations to converge.
Returns:
hemiHemiSphere
Distributed points on a unit sphere.
potentialndarray
The electrostatic potential at each iteration. This can be useful to check if the repulsion converged to a minimum.
Notes
This function is meant to be used with diffusion imaging so antipodal symmetry is assumed. Therefore, each charge must not only be unique, but if there is a charge at +x, there cannot be a charge at -x. These are treated as the same location and because the distance between the two charges will be zero, the result will be unstable.
### disperse_charges_alt
dipy.core.sphere.disperse_charges_alt(init_pointset, iters, tol=0.001)
Reimplementation of disperse_charges making use of scipy.optimize.fmin_slsqp.
Parameters:
init_pointset(N, 3) ndarray
Points on a unit sphere.
itersint
Number of iterations to run.
tolfloat
Tolerance for the optimization.
Returns:
array-like (N, 3)
Distributed points on a unit sphere.
### euler_characteristic_check
dipy.core.sphere.euler_characteristic_check(sphere, chi=2)
Checks the euler characteristic of a sphere
If $$f$$ = number of faces, $$e$$ = number_of_edges and $$v$$ = number of vertices, the Euler formula says $$f-e+v = 2$$ for a mesh on a sphere. More generally, whether $$f -e + v == \chi$$ where $$\chi$$ is the Euler characteristic of the mesh.
• Open chain (track) has $$\chi=1$$
• Closed chain (loop) has $$\chi=0$$
• Disk has $$\chi=1$$
• Sphere has $$\chi=2$$
• HemiSphere has $$\chi=1$$
Parameters:
sphereSphere
A Sphere instance with vertices, edges and faces attributes.
chiint, optional
The Euler characteristic of the mesh to be checked
Returns:
checkbool
True if the mesh has Euler characteristic $$\chi$$
Examples
>>> euler_characteristic_check(unit_octahedron)
True
>>> hemisphere = HemiSphere.from_sphere(unit_icosahedron)
>>> euler_characteristic_check(hemisphere, chi=1)
True
### faces_from_sphere_vertices
dipy.core.sphere.faces_from_sphere_vertices(vertices)
Triangulate a set of vertices on the sphere.
Parameters:
vertices(M, 3) ndarray
XYZ coordinates of vertices on the sphere.
Returns:
faces(N, 3) ndarray
Indices into vertices; forms triangular faces.
### remove_similar_vertices
dipy.core.sphere.remove_similar_vertices(vertices, theta, return_mapping=False, return_index=False)
Remove vertices that are less than theta degrees from any other
Returns vertices that are at least theta degrees from any other vertex. Vertex v and -v are considered the same so if v and -v are both in vertices only one is kept. Also if v and w are both in vertices, w must be separated by theta degrees from both v and -v to be unique.
Parameters:
vertices(N, 3) ndarray
N unit vectors.
thetafloat
The minimum separation between vertices in degrees.
return_mapping{False, True}, optional
If True, return mapping as well as vertices and maybe indices (see below).
return_indices{False, True}, optional
If True, return indices as well as vertices and maybe mapping (see below).
Returns:
unique_vertices(M, 3) ndarray
Vertices sufficiently separated from one another.
mapping(N,) ndarray
For each element vertices[i] ($$i \in 0..N-1$$), the index $$j$$ to a vertex in unique_vertices that is less than theta degrees from vertices[i]. Only returned if return_mapping is True.
indices(N,) ndarray
indices gives the reverse of mapping. For each element unique_vertices[j] ($$j \in 0..M-1$$), the index $$i$$ to a vertex in vertices that is less than theta degrees from unique_vertices[j]. If there is more than one element of vertices that is less than theta degrees from unique_vertices[j], return the first (lowest index) matching value. Only return if return_indices is True.
### sphere2cart
dipy.core.sphere.sphere2cart(r, theta, phi)
Spherical to Cartesian coordinates
This is the standard physics convention where theta is the inclination (polar) angle, and phi is the azimuth angle.
Imagine a sphere with center (0,0,0). Orient it with the z axis running south-north, the y axis running west-east and the x axis from posterior to anterior. theta (the inclination angle) is the angle to rotate from the z-axis (the zenith) around the y-axis, towards the x axis. Thus the rotation is counter-clockwise from the point of view of positive y. phi (azimuth) gives the angle of rotation around the z-axis towards the y axis. The rotation is counter-clockwise from the point of view of positive z.
Equivalently, given a point P on the sphere, with coordinates x, y, z, theta is the angle between P and the z-axis, and phi is the angle between the projection of P onto the XY plane, and the X axis.
Geographical nomenclature designates theta as ‘co-latitude’, and phi as ‘longitude’
Parameters:
rarray_like
thetaarray_like
inclination or polar angle
phiarray_like
azimuth angle
Returns:
xarray
x coordinate(s) in Cartesion space
yarray
y coordinate(s) in Cartesian space
zarray
z coordinate
Notes
See these pages:
for excellent discussion of the many different conventions possible. Here we use the physics conventions, used in the wikipedia page.
Derivations of the formulae are simple. Consider a vector x, y, z of length r (norm of x, y, z). The inclination angle (theta) can be found from: cos(theta) == z / r -> z == r * cos(theta). This gives the hypotenuse of the projection onto the XY plane, which we will call Q. Q == r*sin(theta). Now x / Q == cos(phi) -> x == r * sin(theta) * cos(phi) and so on.
We have deliberately named this function sphere2cart rather than sph2cart to distinguish it from the Matlab function of that name, because the Matlab function uses an unusual convention for the angles that we did not want to replicate. The Matlab function is trivial to implement with the formulae given in the Matlab help.
### unique_edges
dipy.core.sphere.unique_edges(faces, return_mapping=False)
Extract all unique edges from given triangular faces.
Parameters:
faces(N, 3) ndarray
Vertex indices forming triangular faces.
return_mappingbool
If true, a mapping to the edges of each face is returned.
Returns:
edges(N, 2) ndarray
Unique edges.
mapping(N, 3)
For each face, [x, y, z], a mapping to it’s edges [a, b, c].
y
/ / a/
/ / /__________ x c z
### unique_sets
dipy.core.sphere.unique_sets(sets, return_inverse=False)
Remove duplicate sets.
Parameters:
setsarray (N, k)
N sets of size k.
return_inversebool
If True, also returns the indices of unique_sets that can be used to reconstruct sets (the original ordering of each set may not be preserved).
Returns:
unique_setsarray
Unique sets.
inversearray (N,)
The indices to reconstruct sets from unique_sets.
### vector_norm
dipy.core.sphere.vector_norm(vec, axis=-1, keepdims=False)
Return vector Euclidean (L2) norm
Parameters:
vecarray_like
Vectors to norm.
axisint
Axis over which to norm. By default norm over last axis. If axis is None, vec is flattened then normed.
keepdimsbool
If True, the output will have the same number of dimensions as vec, with shape 1 on axis.
Returns:
normarray
Euclidean norms of vectors.
Examples
>>> import numpy as np
>>> vec = [[8, 15, 0], [0, 36, 77]]
>>> vector_norm(vec)
array([ 17., 85.])
>>> vector_norm(vec, keepdims=True)
array([[ 17.],
[ 85.]])
>>> vector_norm(vec, axis=0)
array([ 8., 39., 77.])
### permutations
class dipy.core.sphere_stats.permutations(iterable, r=None)
Bases: object
Return successive r-length permutations of elements in the iterable.
permutations(range(3), 2) –> (0,1), (0,2), (1,0), (1,2), (2,0), (2,1)
__init__(*args, **kwargs)
### angular_similarity
dipy.core.sphere_stats.angular_similarity(S, T)
Computes the cosine distance of the best match between points of two sets of vectors S and T
Parameters:
Sarray, shape (m,d)
Tarray, shape (n,d)
Returns:
max_cosine_distance:float
Examples
>>> import numpy as np
>>> from dipy.core.sphere_stats import angular_similarity
>>> S=np.array([[1,0,0],[0,1,0],[0,0,1]])
>>> T=np.array([[1,0,0],[0,0,1]])
>>> angular_similarity(S,T)
2.0
>>> T=np.array([[0,1,0],[1,0,0],[0,0,1]])
>>> S=np.array([[1,0,0],[0,0,1]])
>>> angular_similarity(S,T)
2.0
>>> S=np.array([[-1,0,0],[0,1,0],[0,0,1]])
>>> T=np.array([[1,0,0],[0,0,-1]])
>>> angular_similarity(S,T)
2.0
>>> T=np.array([[0,1,0],[1,0,0],[0,0,1]])
>>> S=np.array([[1,0,0],[0,1,0],[0,0,1]])
>>> angular_similarity(S,T)
3.0
>>> S=np.array([[0,1,0],[1,0,0],[0,0,1]])
>>> T=np.array([[1,0,0],[0,np.sqrt(2)/2.,np.sqrt(2)/2.],[0,0,1]])
>>> angular_similarity(S,T)
2.7071067811865475
>>> S=np.array([[0,1,0],[1,0,0],[0,0,1]])
>>> T=np.array([[1,0,0]])
>>> angular_similarity(S,T)
1.0
>>> S=np.array([[0,1,0],[1,0,0]])
>>> T=np.array([[0,0,1]])
>>> angular_similarity(S,T)
0.0
>>> S=np.array([[0,1,0],[1,0,0]])
>>> T=np.array([[0,np.sqrt(2)/2.,np.sqrt(2)/2.]])
Now we use print to reduce the precision of of the printed output (so the doctests don’t detect unimportant differences)
>>> print('%.12f' % angular_similarity(S,T))
0.707106781187
>>> S=np.array([[0,1,0]])
>>> T=np.array([[0,np.sqrt(2)/2.,np.sqrt(2)/2.]])
>>> print('%.12f' % angular_similarity(S,T))
0.707106781187
>>> S=np.array([[0,1,0],[0,0,1]])
>>> T=np.array([[0,np.sqrt(2)/2.,np.sqrt(2)/2.]])
>>> print('%.12f' % angular_similarity(S,T))
0.707106781187
### compare_orientation_sets
dipy.core.sphere_stats.compare_orientation_sets(S, T)
Computes the mean cosine distance of the best match between points of two sets of vectors S and T (angular similarity)
Parameters:
Sarray, shape (m,d)
First set of vectors.
Tarray, shape (n,d)
Second set of vectors.
Returns:
max_mean_cosinefloat
Maximum mean cosine distance.
Examples
>>> from dipy.core.sphere_stats import compare_orientation_sets
>>> S=np.array([[1,0,0],[0,1,0],[0,0,1]])
>>> T=np.array([[1,0,0],[0,0,1]])
>>> compare_orientation_sets(S,T)
1.0
>>> T=np.array([[0,1,0],[1,0,0],[0,0,1]])
>>> S=np.array([[1,0,0],[0,0,1]])
>>> compare_orientation_sets(S,T)
1.0
>>> from dipy.core.sphere_stats import compare_orientation_sets
>>> S=np.array([[-1,0,0],[0,1,0],[0,0,1]])
>>> T=np.array([[1,0,0],[0,0,-1]])
>>> compare_orientation_sets(S,T)
1.0
### eigenstats
dipy.core.sphere_stats.eigenstats(points, alpha=0.05)
Principal direction and confidence ellipse
Implements equations in section 6.3.1(ii) of Fisher, Lewis and Embleton, supplemented by equations in section 3.2.5.
Parameters:
pointsarray_like (N,3)
array of points on the sphere of radius 1 in $$\mathbb{R}^3$$
alphareal or None
1 minus the coverage for the confidence ellipsoid, e.g. 0.05 for 95% coverage.
Returns:
centrevector (3,)
centre of ellipsoid
b1vector (2,)
lengths of semi-axes of ellipsoid
### random_uniform_on_sphere
dipy.core.sphere_stats.random_uniform_on_sphere(n=1, coords='xyz')
Random unit vectors from a uniform distribution on the sphere.
Parameters:
nint
Number of random vectors
‘xyz’ for cartesian form ‘radians’ for spherical form in rads ‘degrees’ for spherical form in degrees
Returns:
Xarray, shape (n,3) if coords=’xyz’ or shape (n,2) otherwise
Uniformly distributed vectors on the unit sphere.
Notes
The uniform distribution on the sphere, parameterized by spherical coordinates $$(\theta, \phi)$$, should verify $$\phi\sim U[0,2\pi]$$, while $$z=\cos(\theta)\sim U[-1,1]$$.
References
Examples
>>> from dipy.core.sphere_stats import random_uniform_on_sphere
>>> X.shape == (4, 2)
True
>>> X = random_uniform_on_sphere(4, 'xyz')
>>> X.shape == (4, 3)
True
### HemiSphere
class dipy.core.subdivide_octahedron.HemiSphere(x=None, y=None, z=None, theta=None, phi=None, xyz=None, faces=None, edges=None, tol=1e-05)
Bases: Sphere
Points on the unit sphere.
A HemiSphere is similar to a Sphere but it takes antipodal symmetry into account. Antipodal symmetry means that point v on a HemiSphere is the same as the point -v. Duplicate points are discarded when constructing a HemiSphere (including antipodal duplicates). edges and faces are remapped to the remaining points as closely as possible.
The HemiSphere can be constructed using one of three conventions:
HemiSphere(x, y, z)
HemiSphere(xyz=xyz)
HemiSphere(theta=theta, phi=phi)
Parameters:
x, y, z1-D array_like
Vertices as x-y-z coordinates.
theta, phi1-D array_like
Vertices as spherical coordinates. Theta and phi are the inclination and azimuth angles respectively.
xyz(N, 3) ndarray
Vertices as x-y-z coordinates.
faces(N, 3) ndarray
Indices into vertices that form triangular faces. If unspecified, the faces are computed using a Delaunay triangulation.
edges(N, 2) ndarray
Edges between vertices. If unspecified, the edges are derived from the faces.
tolfloat
Angle in degrees. Vertices that are less than tol degrees apart are treated as duplicates.
Sphere
Attributes:
x
y
z
Methods
Find the index of the vertex in the Sphere closest to the input vector, taking into account antipodal symmetry from_sphere(sphere[, tol]) Create instance from a Sphere Create a full Sphere from a HemiSphere subdivide([n]) Create a more subdivided HemiSphere
edges faces vertices
__init__(x=None, y=None, z=None, theta=None, phi=None, xyz=None, faces=None, edges=None, tol=1e-05)
Create a HemiSphere from points
faces()
find_closest(xyz)
Find the index of the vertex in the Sphere closest to the input vector, taking into account antipodal symmetry
Parameters:
xyzarray-like, 3 elements
A unit vector
Returns:
idxint
The index into the Sphere.vertices array that gives the closest vertex (in angle).
classmethod from_sphere(sphere, tol=1e-05)
Create instance from a Sphere
mirror()
Create a full Sphere from a HemiSphere
subdivide(n=1)
Create a more subdivided HemiSphere
See Sphere.subdivide for full documentation.
### create_unit_hemisphere
dipy.core.subdivide_octahedron.create_unit_hemisphere(recursion_level=2)
Creates a unit sphere by subdividing a unit octahedron, returns half the sphere.
Parameters:
recursion_levelint
Level of subdivision, recursion_level=1 will return an octahedron, anything bigger will return a more subdivided sphere. The sphere will have $$(4^recursion_level+2)/2$$ vertices.
Returns:
HemiSphere
Half of a unit sphere.
### create_unit_sphere
dipy.core.subdivide_octahedron.create_unit_sphere(recursion_level=2)
Creates a unit sphere by subdividing a unit octahedron.
Starts with a unit octahedron and subdivides the faces, projecting the resulting points onto the surface of a unit sphere.
Parameters:
recursion_levelint
Level of subdivision, recursion_level=1 will return an octahedron, anything bigger will return a more subdivided sphere. The sphere will have $$4^recursion_level+2$$ vertices.
Returns:
Sphere
The unit sphere.
create_unit_hemisphere, Sphere
### afb3D
dipy.core.wavelet.afb3D(x, af1, af2=None, af3=None)
3D Analysis Filter Bank
Parameters:
x3D ndarray
N1 by N2 by N3 array matrix, where 1) N1, N2, N3 all even 2) N1 >= 2*len(af1) 3) N2 >= 2*len(af2) 4) N3 >= 2*len(af3)
afi2D ndarray
analysis filters for dimension i afi[:, 1] - lowpass filter afi[:, 2] - highpass filter
Returns:
lo1D array
lowpass subband
hi1D array
highpass subbands, h[d]- d = 1..7
### afb3D_A
dipy.core.wavelet.afb3D_A(x, af, d)
3D Analysis Filter Bank
(along one dimension only)
Parameters:
x3D ndarray
N1xN2xN2 matrix, where min(N1,N2,N3) > 2*length(filter)
(Ni are even)
af2D ndarray
analysis filter for the columns af[:, 1] - lowpass filter af[:, 2] - highpass filter
dint
dimension of filtering (d = 1, 2 or 3)
Returns:
lo1D array
lowpass subbands
hi1D array
highpass subbands
### cshift3D
dipy.core.wavelet.cshift3D(x, m, d)
3D Circular Shift
Parameters:
x3D ndarray
N1 by N2 by N3 array
mint
amount of shift
dint
dimension of shift (d = 1,2,3)
Returns:
y3D ndarray
array x will be shifed by m samples down along dimension d
### dwt3D
dipy.core.wavelet.dwt3D(x, J, af)
3-D Discrete Wavelet Transform
Parameters:
x3D ndarray
N1 x N2 x N3 matrix 1) Ni all even 2) min(Ni) >= 2^(J-1)*length(af)
Jint
number of stages
af2D ndarray
analysis filters
Returns:
wcell array
wavelet coefficients
### idwt3D
dipy.core.wavelet.idwt3D(w, J, sf)
Inverse 3-D Discrete Wavelet Transform
Parameters:
wcell array
wavelet coefficient
Jint
number of stages
sf2D ndarray
synthesis filters
Returns:
y3D ndarray
output array
### permutationinverse
dipy.core.wavelet.permutationinverse(perm)
Function generating inverse of the permutation
Parameters:
perm1D array
Returns:
inverse1D array
permutation inverse of the input
### sfb3D
dipy.core.wavelet.sfb3D(lo, hi, sf1, sf2=None, sf3=None)
3D Synthesis Filter Bank
Parameters:
lo1D array
lowpass subbands
hi1D array
highpass subbands
sfi2D ndarray
synthesis filters for dimension i
Returns:
y3D ndarray
output array
### sfb3D_A
dipy.core.wavelet.sfb3D_A(lo, hi, sf, d)
3D Synthesis Filter Bank
(along single dimension only)
Parameters:
lo1D array
lowpass subbands
hi1D array
highpass subbands
sf2D ndarray
synthesis filters
dint
dimension of filtering
Returns:
y3D ndarray
the N1xN2xN3 matrix | 34,696 | 134,342 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2023-14 | longest | en | 0.546088 |
http://physicsmathforums.com/archive/index.php/t-418.html | 1,371,595,456,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368707435344/warc/CC-MAIN-20130516123035-00067-ip-10-60-113-184.ec2.internal.warc.gz | 198,619,059 | 1,513 | PDA
View Full Version : A question related to fluid dynamics
03-06-2006, 10:51 AM
HELP ME PLZ i m unable to solve the following question
Water flows steadily through the funnel shown in figure. throughout most of the funnelthe flow is approximately radial(along rays from O) with a velocity of V= c/r^2(where cap shows raise to the power),where r is the radial coordinate and c is a constant.if the velocity is 0.4m/s when r=0.1m. determine the acceleration at point A and B.
one more question plz...how could i draw or attach a picture or a figure in this window? | 144 | 568 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2013-20 | latest | en | 0.912548 |
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